Chapter 1 Pythagoras Theorem

Chapter 1

Pythagoras Theorem

1.1 Pythagoras Theorem

Consider a square CDEF partitioned in two diﬀerent ways as shown below. In Figure 1A, the square is decomposed as the square ABP Q together with four congruent triangles. Likewise, in Figure 1B, the same square is decomposed as the squares AKY F , BKXD together with four congruent triangles, each of which is congruent to those in Figure 1A.

P E ✚ D E X D ✚ ❙ ✚ ❙ ✚ ❙ Q✚ ❙ ❙ ❙ K ❙ ✚ B Y ✚ B ✚ ✚ ❙ ✚ ✚ ❙ ✚ ✚ F ❙✚ C F ✚ C A A Note that the square AKY F (respectively BKXD) has the same area as the square constructed on the side BC (respectively AC)oftheright triangle ABC. Comparing the two ﬁgures, we conclude that for the right triangle ABC, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two shorter sides.

This is the famous Pythagoras Theorem: The lengths a, b, c of the sides of a right triangle satisfy the relation a2 + b2 = c2. Here, of course, c is the length of the hypotenuse,thelongest among the three sides.

1 YIU : Trigonometry Workbook 2

Exercises 1. In each of the following right triangles, two of the sides are given. Find the length of the remaining side.

a b c (i) 3 4 (ii) 13 14 (iii) 20 29 (iv) 19 29 (v) 60 61 (vi) 59 61

2. A ladder 20 feet long is leaning against a wall. The end on the ground is 12 feet from the wall. How high up the wall does the ladder reach ?

3. The lengths of the three sides of a right triangle are x, x +1,andx +2.Findx.

4. The lengths of the three sides of a right triangle are 1, x,andx2.Findx. YIU : Trigonometry Workbook 3

5. A man who wants to carry a long stick of unknown length vertically through a door ﬁnds that the stick is 2 feet too long. If he carries it horizontally, it is 4 feet too long. However, it just makes diagonally. How long is the stick, and what are the dimensions of the door ?

6. Find the side of a rhombus whose diagonals are 10 and 24 cm.

7. In a circle of radius 65 cm there is a chord 66 cm. How far is the chord from the center ?

8. Each side of an equilateral triangle has length a inches. Find the area of the triangle. YIU : Trigonometry Workbook 4

9. Solve the following right triangles. Given a b c (i) a =40,c+ b =50 ∗ (ii) a =35,c− b =25 ∗ (iii) c =29,c+ b =50 ∗ (iv) a =45,c− b =25 ∗ (v) c + a =25,c+ b =32 (vi) c + a =81,c− b =2 (vii) c − a =25,c− b =32

Hints and Comments 2. Answer: 16 ft. 3. x = 3. The other√ two sides are 4 and 5. 1 4. x = 2 ( 2+2 5). 5. x = 10. The dimensions of the door are 6 ft. and 8 ft. 6. A rhombus is a quadrilateral whose 4 sides are equal in length. As such, it must be a parallelogram. Morever, the diagonals (of a rhombus) are perpendicular to each other. Answer: 13 cm. 7. This means ﬁnding the length of the perpendicular from the center to the chord. You need to know that this perpendicular passes through the midpoint of the chord. Answer: 56 cm.

8. The perpendicular from a vertex√ (of an equilateral triangle) to its opposite side passes through 3 2 the midpoint of this side. Answer: 4 a . 9. (i) b =9,c= 41; (ii) b =12,c= 37; (iii) a =20,b= 21; (iv) b =28,c= 53; (v) a =8,b=15,c= 17; (vi) a =16,b=63,c= 65; (vii) a =72,b=65,c= 97.

Given abcRemark a c ± b 1 c ± b a2 1 c ± b − a2 (i), (ii) , * 2 (( )+ c±b ) 2 (( ) c±b ) (iii) c, c + b c2 − [(c + b) − c]2 (c + b) − c * 1 a2 1 a2 (iv) a, c − b * 2 ((c − b) − c−b ) 2 ((c − b)+ c−b ) (v) c + a, c + b 2(c + a)(c + b)=(a + b + c)2 (vi) c + a, c − b 2(c + a)(c − b)=(a − b + c)2 (vii) c − a, c − b 2(c − a)(c − b)=(a + b − c)2 YIU : Trigonometry Workbook 5

1.2 *Right triangles with integer sides

1. Take any two numbers m and n. Assuming m>n,formthenumbers

a = m2 − n2,b=2mn, c = m2 + n2.

Show that a, b and c are the lengths of the sides of a right triangle.

2. It is known that if the three sides of a right triangle are all integers and do not have common divisors, then these three sides can be obtained by the method in Exercise 1 by suitably choosing integers m and n, one odd, one even, and not having common divisors. Complete the following table to ﬁnd all such right triangles with sides not exceeding 100:

mna= m2 − n2 b =2mn c = m2 + n2 (i) 2 1 3 4 5 (ii) 3 2 (iii) 4 1 (iv) 4 3 (v) 5 2 (vi) 5 4 (vii) 6 1 (viii) 6 5 mna= m2 − n2 b =2mn c = m2 + n2 (ix) 7 2 (x) 7 4 (xi) 7 6 (xii) 8 1 (xiii) 8 3 (xiv) 8 5 (xv) 9 2 (xvi) 9 4 65 72 97

a ≥ b 1 a2 − c 1 a2 3. Let 3beanoddnumber.Form = 2 ( 1) and = 2 ( + 1). Showthat (a) c = b +1; (b) a, b, c form the sides of a right triangle. YIU : Trigonometry Workbook 6

Indeed, this is a description of all integer right triangles whose two longer sides diﬀer by 1. Complete the following tables to list the ﬁrst 10 such triangles.

a 3579111315171921 b 412 c 513

4. Howabout integer right triangles with the two shorter sides differing by 1 ? The first two examples are (3,4,5) and (20,21,29) as you can see from the table in Exercise 3. The others can be generated from these two as follows. Define a sequence of integers (xn) recursively by

xn := 6xn−1 − xn−2 +2; x1 =3,x2 =20.

(a) Complete the following list of xn for n =3, 4,...,10. 2 2 (b) For each n,verifyxn +(xn +1) isthesquareofanintegeryn.

k 12345678910 xn 320 yn 529

Hints and Comments 1. The converse of Pythagoras Theorem is true: if the square on the longest side of a triangle is equal to the sum of the squares on the remaining two sides, then the angle opposite to the longest side is a right angle. See Exercise 2.2.10 for an explanation. Simply verify that (m2 −n2)2 +(2mn)2 =(m2 +n2)2. 2**. There are many explanations of this fact. Here is one. If a, b,andc are integers satisfying 2 2 2 a b 2 2 a +b = c , then writing x = c and y = c ,wehavex +y =1.Notethatx and y are rational numbers. On the other hand, if x and y are rationalnumbers such that x2 + y2 = 1, then writing x and y as a b 2 2 2 fractions with a common denominator, say, s = c and y = c ,wehavea + b = c .Itfollowsthatright triangles with integer sides correspond to rational points in the ﬁrst quadrant on the unit circle.Now, using coordinate geometry, it is possible to describe all rationalpoints on the circle x2 + y2 =1. First of all, there is the point A =(−1, 0) on this circle. If P is a point with rational coordinates, then the slope of the line AP must always have rational coordinates. It follows that every rational point on the circle can be obtained as the intersection of the circle with a line through A with rational slope. Now, every such line has equation y = k(x + 1) for a rationalnumber k.Sincex2 + y2 =1,wehave

x2 + k2(x +1)2 =1, (1 + k2)x2 +2k2x − (1 − k2)=0, (x + 1)[((1 + k2)x − (1 − k2)] = 0, 1 − k2 x = −1, . 1+k2 Correspondingly, from y = k(x +1),wehave

2k y =0, . 1+k2 YIU : Trigonometry Workbook 7

2 2 1−k2 2k This shows that every rational point on the circle x + y =1is of the form ( 1+k2 , 1+k2 ). Now, writing, n k = m for integers m and n which do not share common divisors, we have

1 − ( n )2 m2 − n2 x m = n 2 = 2 2 ; 1+(m ) m + n 2 · n mn y m 2 . = n 2 = 2 2 1+(m ) m + n

It follows that if a, b, c do not have common divisor, and are the sides of an integer triangle, then with suitable choice of m and n,

a = m2 − n2,b=2mn, c = m2 + n2.

Note that m and n are of diﬀerent parity, i.e., one even, one odd, for otherwise, a, b and c would be all even, contrary to the assumption that they do not share common divisors. 2 1 2 1 2 3. If c − b =1,thenc + b = a .Itfollowsthatc = 2 (a +1)andb = 2 (a − 1). Now, a must be an odd number. If we write a =2k +1,thenb =2k(k +1),andc =2k2 +2k +1.

a 3 5 7 9 11 13 15 17 19 21 b 4 12 24 40 60 84 112 144 180 220 c 5 13 25 41 61 85 113 145 181 221

4. xn 3 20 119 696 4059 23660 137903 803760 4684659 27304196 xn + 1 4 21 120 697 4060 23661 137904 803761 4684660 27304197 yn 5 29 169 985 5741 33461 195025 1136689 6625109 38613965

Indeed, yn can be deﬁned recursively as follows.

yn := 6yn−1 − yn−2; y1 =5,y2 =29. Chapter 2

Congruence and similarity

2.1 Construction of triangles

Notation Let A, B and C denote the vertices of a triangle. Each vertex has an opposite side. The length of the side BC opposite to A is denoted by a; likewise, b and c denote respectively the lengths of the sides opposite to B and C. The triangle has three angles,eachmeasuredindegrees.Themeasure of the angle at A (respectively B, C)is also denoted by the same symbol A (respectively B and C).

Fundamental Principles of Euclidean Geometry (A) Each external angle of a triangle is equal to the sum of the measures of the remaining two internal angles. (B) The sum of the measures of the three angles of a triangle is always 180◦. (C) Alternate angles between parallel lines are equal.

A x❅ ❅ a ❅ ❅ ❅ ❅ ❅ b y ❅ z B C x + y = z a = b

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Exercises In each of the following exercises, try to construct a triangle with three measurements given. Speciﬁcally note if the answer is unique. 1(a). a =4;b =5;c =6.

1(b). What condition must a, b and c satisfy so that they there is a triangle with sides a, b,andc ?

2(a). a =4,b =5,C =60◦. YIU : Trigonometry Workbook 10

2(b). a =5,b =8,A =30◦.

2(c). a =3,b =8,A =30◦.

2(d). a =4,b =8,A =30◦. YIU : Trigonometry Workbook 11

3. a =4,B =40◦, C =60◦.

4. A =50◦, B =60◦.(Ofcourse,C must be 70◦).

Hints and Comments 1(b) Triangle inequality: the largest of the three must be smaller than the sum of the other two. 2(b) 2 triangles; 2(c) unique; 2(d) not possible. 3. Unique. 4. The three angles determine the shape of the triangle, but not the size. YIU : Trigonometry Workbook 12

2.2 Congruence of triangles

Two triangles are congruent if it is possible to set up a correspondence of the vertices so that corresponding angles are equal in measure and corresponding sides are equal in length. ABC ≡XYZ means

A = X, B = Y, C = Z; AB = XY, BC = YZ, CA= ZX.

A X ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ BCZY Congruence tests for triangles

SSS three pairs of equal sides SAS two pairs of equal sides and equal included angles AAS or ASA two pairs of equal angles and one pair of equal corresponding sides RHS right triangles with equal hypotenuses and one more pair of equal sides

Exercises Use congruence tests for triangles to explain each of the following items. 1. The base angles of an isosceles angle are equal. Why does it follow that each angle of an equilateral triangle is 60◦ ? YIU : Trigonometry Workbook 13

2(a). The line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.

2(b). The perpendicular from the center of a circle to a chord passes through the midpoint of the chord.

3(a). The diagonals of a parallelogram bisect each other.

3(b). The two diagonals of a rectangle are equal in length. YIU : Trigonometry Workbook 14

3(c). The two diagonals of a rhombus are perpendicular to each other.

4. Let X, Y and Z be the midpoints of the sides BC, CA and AB of ABC respectively. The triangles XY Z, AZY , ZBX,andYXC are congruent.

5 (Perpendicular bisector locus) Let A and B be ﬁxed points. A point P is equidistant from A and B if and only if P lies on the perpendicular bisector of the segment AB. Proof. (Necessity) Assume AP = BP.

(Suﬃciency) Let P be a point on the perpendicular bisector of AB. YIU : Trigonometry Workbook 15

6. (Angle bisector locus) Let l and l be two intersecting lines. A point P is equidistant from l and l if and only if P lies on the bisector of one of the angles between the lines.

7. Let A be a point outside a given circle C. There are two tangents from A to C.If P and Q are the points of contact of these tangents with the circle, then AP = AQ.

8. Let AB be a diameter of a circle, and P any point on the circle other than these two. Show that AP B =90◦. YIU : Trigonometry Workbook 16

9. Let A and B be two points on a circle C, center O. Assume that AB is not a diameter of the circle, so that the points divide the circle into a major arc and a minor arc. (a) Let P be a point on the major arc AB, i.e., P and the center O are on the same side of the line AB. Show that AOB =2 AP B.

(b) If Q is a point on the minor arc AB, i.e. Q and O are on opposite sides of the line AB, show that AOB +2 AQB = 360◦.

10. (Converse of Pythagoras Theorem) Suppose a2 +b2 = c2 in ABC.Consider another triangle ABC with BC = a, CA = b and C =90◦. Compute the length of AB and explain why ABC and ABC are congruent. Conclude from this that C =90◦. YIU : Trigonometry Workbook 17

Hints and Comments 1. Suppose AB = AC.ThenABC ≡ACB by the SSS test. It follows that B = C.(If you really want to see two triangles, you may mark the midpoint M of the side BC and explain why ABM ≡ ACM). 2(a). If M is the midpoint of a chord AB of a circle, center O, OAM ≡ OBM by the SSS test. It follows that the two angles at M are equal, and OM ⊥ AB. 2(b). Let P be a point on a chord AB of a circle, center O. OAP ≡ OBP by the RHS test. It follows that AP = BP. 3(a). Let ABCD be a parallelogram whose diagonals meet at M. AMB ≡ CMD by the ASA test. It follows that AM = CM and BM = DM. 3(b) If ABCD is a rectangle, ABC ≡ BAD by the SAS test. It follows that AC = BD. 3(c) Let ABCD be a rhombus whose diagonals meet at M. Since these diagonals bisect each other, ABM ≡ ADM by the SSS test. Thus, AMB = AMD and AM ⊥ BD. 1 4. If you can see that XY = 2 AB = AZ = BZ, and similarly for YZ and ZX, the congruence of these triangles follows easily from the SSS test. 6. The distance from a point to a line is the ‘perpendicular distance’. Let P be a point outside a given line l,andQ a point on l such that PQ ⊥ l. Then the length PQ is the distance from P to l. 8. This means that if you want to draw a circle passing through the vertices of a right triangle, you should take the midpoint of the hypotenuse as the center. 9. The converses of (a) and (c) are also true. Let A, B, P and Q be points in a plane. (i) If P and Q are on the same side of AB,andif AP B = AQB, then the four points lie on a circle. (ii) If P and Q are on opposite sides of AB,andif AP B + AQB = 180◦, then the four points lie on a circle. YIU : Trigonometry Workbook 18

2.3 Similarity of triangles

Two triangles are similar if there is a correspondence of their vertices such that corresponding angles are equal and pairs of corresponding sides are in proportion.

A ❅ ❅ ❅ ❅ X ❅ ❙ ❅ ❙ ❅ ❙ ❅ ❙ BCZY

ABC XY Z means AB BC CA A X, B Y, C Z . = = = ; XY = YZ = ZX

Similarity tests Two triangles are similar if they contain 1. two pairs of equal corresponding angles, or 2. two pairs of corresponding sides in proportion, and with equal included angles, or 3. three pairs of corresponding sides in proportion.

Exercises 1. (The Midpoint Theorem) Let X and Y be the midpoints of the sides BC and CA of ABC. Explain why YXC and ABC are similar. Conclude from this that XY 1 AB = 2 . YIU : Trigonometry Workbook 19

2. (An alternative proof of Pythagoras Theorem) Let C be the vertex of the right angle of ABC,andP the point on the hypotenuse AB such that CP ⊥ AB.

(a) Show that ACP ABC by identifying two pairs of equal angles. Conclude from this that AC2 = AP · AB.

(b) Show that CBP ABC by identifying two pairs of equal angles. Conclude from this that BC2 = BP · AB. YIU : Trigonometry Workbook 20

(d) Why are ACP and CBP similar ? Conclude from this that CP2 = AP · BP.

3. Let P be a point on the hypotenuse AB of right triangle ABC,andX, Y its projections on the sides BC and CA respectively. If CXPY is a square, ﬁnd the length of each side of this square.

4. (Theorem of intersecting chords) Let AB and CD be two chords of a circle intersecting at a point P , say, inside the circle. (a) Why are ACP and DBP similar ? Conclude from this that AP ·PB = CP·PD. YIU : Trigonometry Workbook 21

(b) Repeat part (a) by assuming that the point P lies outside the circle.

5. Let X and Y be the projections of two points A and B onthesamesideofaline l. The lines AY and BX meet at a point P .IfQ is the projection of P on l, show that

1 1 1 . AX + BY = PQ

Hints and Comments 2(d) Of course you can show this directly. But if you realize that the similarity relation of triangles is transitive, you could have omitted all the details. Note the this similarity relation is also reﬂexive and symmetric.Thisisanexampleofanequivalence relation, so is the congruence of triangles. a−x a ab 3. PBX ABC. If each side of the square has length x,then x = b ; x = a+b . 4. Let P be a point inside a circle, radius r.DrawachordAB of the circle through P .LetM be the midpoint of AB. Exercise: Use Pythagoras Theorem to show that AP · BP = r2 − OP 2.How should this relation be modiﬁed if P lies outside the circle. How does this give an alternative proof of the theorem of intersecting chords ? Chapter 3

Areas

3.1 Basic area principles

Basic area principles (A) The area of a rectangle of length a units and width b units is ab square units. (B) Suppose two areas are intercepted between two parallel l1 and l2. If each line parallel to l1 and l2 intercepts segments of equal lengths in these two areas, then the two areas are equal. 1 × × (C) Area of triangle = 2 base height.

Exercise 1. The diagonal of a rhombus has length 12 cm. and 15 cm. What is its area ?

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2 Is it possible to construct a triangle whose altitudes are 1, 2 and 3 ?

3. The heights of a parallelogram are h and k, and its perimeter is p.Findthearea of the parallelogram.

4. In ABC, suppose the bisector of BAC meets the side BC at X. (a) By considering ABX and ACX as triangles with a common altitude from A BC ABX BX to , show that ACX = CX .

ABX AB (b) Show also that ACX = AC , and deduce that in a triangle, each angle bisector divides the opposite side into the ratio of the remaining sides. YIU : Trigonometry Workbook 24

(c) Suppose AB > AC.IfX is a point on the extension of BC such that AX bisects the external angle of A,then BX AB = . CX AC

5. Let A and B be two ﬁxed points at a distance 6 cm. apart, and P a variable point such that AP : BP =2:1. (a) Suppose the bisector of AP B meets AB at X.CalculateBX.

(b) Suppose the external bisector of AP B meets AB at X.CalculateBX.

6*. (Euclid’s proof of Pythagoras Theorem) Let ABC be a right triangle with C =90◦.Consider (i) the square ACHK on the side AB; (ii) the square ABP Q on the hypotenuse AB; (iii) the perpendicular from C to AB, meeting AB at X and PQ at Y .

(a) Show that ACHK =2ABK,andAXY Q =2AQC.

(b) Why do ABK and AQC have the same area ? Deduce that ACHK = AXY Q.

Hints and Comments YIU : Trigonometry Workbook 26

1. Answer: 90 sq. cm. Since the diagonals of a rhombus are perpendicular to each other, the area is one half of the product of the lengths of the diagonals. 2. An altitude of a triangle is the (length of the) segment from a vertex to its opposite side. If there 2 2 2 is a triangle with altitudes 1,2,3, and area , then the lengths of the sides are 1 , 2 ,and 3 .These 1 1 are in the ratio 1 : 2 : 3 =6:3:2.Since6< 3 + 2, this contradicts the triangle inequality.Nosuch triangle exists. T T T T 3. If the area is T , then the lengths of the sides are h and k , and the perimeter is 2( h + k )=p. hk From this, T = (2h+k) p. 5. (a) 2 cm. (b) 6 cm. YIU : Trigonometry Workbook 27

3.2 Heron formula for the area of a triangle

Consider a triangle ABC with lengths of sides a, b, c. Inside the triangle, there is a circle tangent to all three sides of the triangle. This is the incircle of the triangle.

To justify the existence of such a circle, we try to locate its center I.Ifsuchan incircle touches the sides BC, CA,andAB respectively at X, Y ,andZ, then clearly, IX = IY = IZ,theradius of the circle. This means that I must lie on the bisector of each of the angles.

1. Why should these three bisectors be meeting at one point ?

Indeed, if we extend the bisector AI and construct the bisectors of the external angles at B and C, these three lines again intersect at a point, say I.WithI as center, we can construct another circle, tangent to BC,andtheextensionsofAB and AC. YIU : Trigonometry Workbook 28

Now, we proceed to determine the radii of these incircle and excircle.

Notation: r = inradius, radius of the incircle; r = exradius for A, the radius of the excircle on the opposite side of A. = area of the triangle.

1 a b c r 2. Show that = 2 ( + + ) .

Denote by X, Y and Z the points of contact of this incircle with AY = AZ, BZ = BX, and CX = CY . If the lengths of these pairs are x, y, z respectively, then y + z = a, x ++z = b, x + y = c. 3. Solve these equations to obtain 1 1 1 x = (b + c − a),y= (c + a − b),z= (a + b − c). 2 2 2

s 1 a b c Here, it is convenient to introduce the semiperimeter := 2 ( + + ) of the triangle, YIU : Trigonometry Workbook 29 which helps make the expressions for x, y and z simpler: 4. Show that (a) x = s − a, y = s − b, z = s − c.

(b) = rs.

Consider the projections P and Q of the excenter I on the sides AB and AC respectively. 5. Show that AP = AQ = s,andthatBP = s − c.

AIZ AIP r s−a 6. Why are triangles and similar ? Conclude from this that r = s .

7. Why are triangles BIZ and IBP similar ? Conclude from this that r · r = (s − b)(s − c). YIU : Trigonometry Workbook 30

8. Determine r and r in terms of s, s − a, s − b,ands − c.

9. Make use of the simple relation = rs (see Exercise 4(b)) to deduce the Heron formula = s(s − a)(s − b)(s − c).

10. Use the Heron formula to calculate the area of the triangles with given sides.

abcss− as− bs− c r = s (i) 3 4 5 6 3 2 1 6 1 (ii) 4 5 6 √ √ 1 (iii) 4 5 7 8 4 3 1 4 6 2 6 (iv) 5 6 7 (v) 5 12 13 (vi) 6 8 10 (vii) 6 25 29 (viii) 7 15 20 (ix) 9 10 17 (x) 13 14 15 YIU : Trigonometry Workbook 31

11. The altitudes a triangle are 12, 15 and 20. What is the area of the triangle ?

12*. (3,4,5) and (13,14,15) are examples of triangles whose sides are consecutive integers and whose areas are also integers. These are the two smallest triangles with this property. In general, if b−1, b and b+1 are three consecutive integers forming the sides of a triangle with integer area, then b must be a member of the sequence deﬁned inductively as follows. bn := 4bn−1 − bn−2; b1 =4,b2 =14. (a) Complete the following table for the next 5 triangles with this property.

n 1234567 bn 414 bn − 13 bn +1 5 n 6

(b) The areas of these triangles can also be computed recursively by

n =14n−1 −n−2; 1 =6, 2 =84.

Verify this for the 3,...,7.

Hints and Comments YIU : Trigonometry Workbook 32

10. The ﬁve triangles in (v) to (ix) are the only ones with integer sides and numerically equal area and perimeter. 2 2 2 11. Let denote the area. The lengths of the sides are 12 , 15 ,and 20 . The semiperimeter is 1 1 1 1 ( 12 + 15 + 20 ) = 5 .BytheHeronformula 1 1 1 1 2 1 1 1 1 1 1 1 = 2 ( − )( − )( − )=2 · · · = 2 · . 5 5 6 5 15 5 10 5 30 15 10 150 From this, = 150. The lengths of the sides are 25, 20 and 15. More generally, if the altitudes are h, k and l, the area of the triangle is given by 1 1 1 1 − 1 1 1 1 − 1 1 1 1 − 1 = (h + k + l )( h + k + l )(h k + l )(h + k l ) YIU : Trigonometry Workbook 33

3.3 Area and circumference of circles

You must have been familiar with the formula for the area of a circle, radius r:

Area = πr2, π 22 . where is a constant which for most practical purposes is taken to be 7 or 3 14. You should try not to replace π by these approximations too hastily. An explanation of this formula is given in the next section.

Exercise 1. The centers of four identical circles each of radius r are at the corners of a unit square. Suppose each circle is tangent to its two neighbours, ﬁnd the area bounded by the circles.

2. Three identical circles each of radius r are mutually tangent to each other externally. Find the area bounded by the circles.

Circumference of a circle. Archimedes has proved that the area of a circle is equal to that of the triangle with height equal to the radius and base equal to the circumference of the circle. 3. Make use of this to show that the circumference of a circle of radius r is 2πr. YIU : Trigonometry Workbook 34

3.4 * On the number π

The basic idea of ﬁnding the area of a circle is to approximate the circle by polygons. A good starting point is the inscribed regular hexagon. 1. Let P1 and P2 be two points on the circle such that P1P2 = r.Forn =3, 4, 5, 6, 7, let Pn be the point on the circle which is diﬀerent from Pn−2 and has distance r from Pn−1. (a) Explain why P7 is the same point as P1.

(b) Why is P1P2P3P4P5P6 a regular hexagon ? What is the area of this hexagon.

2. Let B be the midpoint of the minor arc P1P2,andK the intersection of OB and P1P2. (a) Why is BP1 one side of a regular 12–gon inscribed in the circle ? YIU : Trigonometry Workbook 35

2 (b) Let A2 denote the area of such a regular 12–gon. Explain why A2 =3r .

Beginning with the inscribed regular hexagon, by repeated bisections of the minor arcs between adjacent vertices of inscribed regular polygons, we obtain sucessively inscribed k regularpolygonsof12,24,48,..., 6· 2 sides, for every positive integer k.Letbk be the k length of one side of a inscribed regular polygon of 6 · 2 sides. Note that b0 =1. b2 r r − r2 − b2 3. Show that k+1 = (2 4 k). YIU : Trigonometry Workbook 36

2 2 4. Let ck := 4r − bk. Show that c0 =3r,and √ ck+1 = r(2r + ck).

5. Deduce that √ 2 ck = 2+ 2+···+ 2+ 3 r , (k fold square root).

6. Consider two adjacent vertices A and B of an inscribed regular polygon of 6 · 2k sides. Let P be the midpoint of the minor arc AB. (a) Why is AP a side of an inscribed regular polygon of 6 · 2k+1 sides ? YIU : Trigonometry Workbook 37

k+1 k (b) Show that the area of an inscribed polygon of 6 · 2 sides is Ak+1 =3· 2 rbk.

7. Use the relations in Exercises 5 and 6 to show that √ k 2 Ak+1 =3· 2 2 − 2+···+ 2+ 3r , (k − 1 fold square root).

√ k 2 Let πk := 3 · 2 2 − 2+···+ 2+ 3r ,(k − 1 fold square root) so that the area k 2 of the inscribed regular 6 · 2 −gon is Ak = πkr . YIU : Trigonometry Workbook 38

k kn=6· 2 Ak Ak+1 2Ak+1 − Ak 1 6 2.598076211r2 3r2 3.401923789r2 2123r2 3.105828541r2 3.211657082r2 3 24 3.105828541r2 3.132628613r2 3.159428685r2 4 48 3.132628613r2 3.139350203r2 3.146071793r2 5 96 3.139350203r2 3.141031951r2 3.142713699r2 This table suggests that 6 192 3.141031951r2 3.141452472r2 3.141872994r2 7 384 3.141452472r2 3.141557608r2 3.141662744r2 8 768 3.141557608r2 3.141583892r2 3.141610176r2 9 1536 3.141583892r2 3.141590463r2 3.141597034r2 10 3072 3.141590463r2 3.141592106r2 3.141593749r2 as n increases, the value of πn gets closer and closer to a certain number, which we denote by π. In other words, if we approximate the circle by an inscribed regular n−gon, for values of n large enough, the area of the regular n−gon is close to πr2.Inthisway,we say that the area of the circle is πr2. For practical calculations, we shall simply take 22 π =3.142 or or 3.1415926 7 from your calculator. What is really signiﬁcant is the existence of such a constant from which one can determine the area of a circle.

Hints and Comments 1(a) The ﬁve triangles OP1P2, OP2P3, OP3P4, OP4P5, OP5P6 are all equilateral. Each of the angles ◦ ◦ ◦ ◦ P1OP2, P2OP3, P3OP4, P4OP5, P5OP6 is 60 .Itfollowsthat P6OP1 = 360 − 5 × 60 =60. P6OP1 is also equilateral, and P6P1 = r. This means that the point P7 coincides with P1. r ◦ (b) Each of the sides√ has length√ and each of its angles is 120 . 1 3 3 3 2 (c) A6 =6× 2 r × 2 r = 2 r . ◦ 1 ◦ 2(a) POB =30. In general, if A and B are two points on a circle such that AOB = n · 360 , then AB can be taken as a side of a regular n−gon inscribed in the circle. 1 P1P2 1 2 (b) Note that OB is perpendicular to and bisects P1P2. The area of BOP1 = 2 ×OB× 2 = 4 r . 2 It follows that A12 =12BOP =3r . k (c) Clearly, if Ak is the area of an inscribed regular polygon of 6 · 2 sides, then circle, Ak

2 2 2 2 2 5. Since n an(r − an)=An, A2 a4 − r2a2 − n , n n = n2 A2 a4 − r2a2 −4 n , 4 n 4 n = n2 n2r4 − A2 a4 − r2a2 r4 4 n 4 n 4 n + = n2 2 4 2 2 2 2 n r − 4An (2an − r ) = n2 2 4 2 1 n r − 4An a2 = r2 − n 2 n

From this, n2r4 − A2 1 2 4 n A2n = nran = nr r − . 2 n YIU : Trigonometry Workbook 40

3.5 Radian measure of an angle

The length of a circular sector is proportional to the angle of at the center. The angles of circular sectors of length equal to the radius of the circle are all equal and are deﬁned to be one radian. It follow that the complete circle can be regarded as a sector with angle 2π radians. n◦ n π 1. Show that an angle of is equal to 180 radians.

x · x 2. Show that an angle of radians is equal to 180 π degrees.

3. Complete the following table of conversion between degree and radian measures of angles.

Degrees 30◦ 45◦ 60◦ 90◦ π 2π 3π π Radians 3 2 2

4. A circular sector of radius r has angle θ radians. Show that (i) the length of the circular arc is rθ, 1 r2θ (ii) the area of the sector is 2 . Chapter 4

Trigonometric functions of acute angles

4.1 The sine and cosine functions

The problem of solution of a right triangle is to completely determine the angles and sides of the triangle given two measurements other than the right angle. 1. If any two of sides are given, the third side can be determined from Pythagoras Theorem. How about the angles ? Try this for (a) c =2anda =1;(b)a = b =1.

If one of the acute angles is known, the remaining angle can be found without eﬀort. How about the sides ? They are only determined up to proportionality. Any two right triangles with a pair of equal acute angles are similar. We reformulate this fact to deﬁne the trigonometric functions of an acute angle which help solve the problem of “solution of right triangles” completely. Let θ be an acute angle. Any two right triangles containing an acute angle equal to θ are similar. Let ABC and ABC be right triangles such that A = A = θ,and

41 YIU : Trigonometry Workbook 42

C = C =90◦.ThenABC ABC, and we have AB BC CA = = . AB BC CA The first equation can be rewritten as BC BC , AB = AB and this fact can be construed as saying that in any right triangle containing an acute θ, the ratio of the opposite side of this angle to hypotenuse does not depend on the size of the triangle. It depends only the shape, and can be regarded as depending only on the angle θ. This leads to the definition of the sine function side opposite to θ sin θ := hypotenuse in any right triangle containing θ as an angle. Similarly, we define the cosine and the tangent functions by side adjacent to θ cos θ := , hypotenuse side opposite to θ tan θ := side adjacent to θ in any right triangle containing θ as an angle.

Trigonometric version of Pythagoras Theorem For any acute angle θ, sin2 θ +cos2 θ =1. θ sin θ 2. Show that tan = cos θ .

3. Solve the right triangle with C =90◦ and given values of YIU : Trigonometry Workbook 43

(a) A and a;

(b) A and b;

4. For most acute angles, it is impossible to compute the precise values of their trigonometric functions directly from the above deﬁnition. In practice, we make use of tables or calculators. Here are some special angles whose trigonometric functions can be easily determined.

θ degrees sin θ cos θ tan θ π ◦ 1 (i) 6 30 2 π ◦ (ii) 3 60 π ◦ (iii) 4 45 1 (iv) 0 0◦ π ◦ (v) 2 90

5. Explain why, for acute angles, the sine function is increasing and the cosine function is decreasing. YIU : Trigonometry Workbook 44

6. The two acute angles of a right triangle are said to be complementary to each other. (This is the same as saying that they add up to 90◦). Explain why thesinefunction of an angle is equal to the cosine function of its complement:

sin θ =cos(90◦ − θ).

7. Complete the following tables (without directly ﬁnding the measures of the acute angles).

sin θ cos θ tan θ 1 (i) 3 1 (ii) 4 1 (iii) 2

8. Find the following trigonometric functions using calculator or mathematical tables. Give your answers to 3 decimal places.

θ degrees sin θ cos θ tan θ π ◦ (i) 18 10 π ◦ (ii) 9 20 2π ◦ (iii) 9 40 5π ◦ (iv) 18 50 π ◦ (v) 10 18 π ◦ (vi) 5 36 3π ◦ (vii) 10 54 2π ◦ (vii) 5 72

Hints and Comments 1(a) If you take a congruent copy of the triangle and glue the two together along the side b, you get an equilateral triangle. Why ? From this it is easy to see that A =30◦ and B =60◦. YIU : Trigonometry Workbook 45

(b) A = B =45◦. 4(iv) Of course there is no right triangle with an angle 0◦. You might want to think of a right triangle with a very small acute angle. 5. Consider two points P and Q on a quadrant of a circle, with center O,andhorizontal radius OA. If AOP > AOQ,thenP is higher than Q.SinceAP = AQ,sinAOP > sin AOQ. The graphs the sine, cosine, and tangent functions in the range 0 ≤ θ ≤ 90◦ are shown in the next two pages. YIU : Trigonometry Workbook 46

4.2 * Trigonometric functions of special angles

1. (The 15◦ family) Let ABC be an equilateral triangle each side of length 2. (i) Draw the minor circular arc with A as center, passing through the points B and C. (ii) Let M be the midpoint of BC.JoinAM and extend to meet the minor arc in (i) at P . √ √ PBM ◦ ◦ − 1 − 2 (a) Explain why =15, and show that tan 15 =2 3= 2 ( 3 1) .

√ ◦ (b)√ Compute sin 15√ .Notethatin√ PBM, BE2 = BD2+DE2 =1+(2−√ 3)√2 = ···= √4(2 − √3) = 2(4 − 2 3) = 2( 3 − 1)2.(exercise). It follows that BE = 2( 3 − 1) = 6 − 2. From this, √ √ √ √ 1 ( 3 − 1)2 3 − 1 6 − 2 sin 15◦ = √2 √ = √ = . 2( 3 − 1) 2 2 4

What is cos 15◦ ?

θ degrees sin θ cos θ tan θ π ◦ (i) 12 15 5π ◦ (ii) 12 75 YIU : Trigonometry Workbook 47

2. Complete the following tables. θ degrees sin θ sin 2θ sin 2θ =2sinθ ? π ◦ (i) 6 30 (a) π ◦ (ii) 4 45 π ◦ (iii) 12 15

θ degrees tan θ tan 2θ tan 2θ =2tanθ ? π ◦ (i) 6 30 (b) π ◦ (ii) 4 45 π ◦ (iii) 12 15

3. (The trigonometric functions of the 18◦ family)LetABC be isosceles with A =36◦, B = C =72◦, a = 2. Let the bisector of B intersect AC at D. (a) Show that BD = AD =2.

(b) Suppose CD = t. Show that ◦ B 1 ABC (i) cos 72 =cos = t+2 in ; YIU : Trigonometry Workbook 48

◦ C t BCD (ii) cos 72 =cos = 4 in .

√ √ t − ◦ 1 − (c) Deduce that = 5 1andcos72 = 4 ( 5 1).

√ ◦ 1 BC ◦ t ◦ (d) Deduce that cos 36 = 4 ( 5 + 1) from the relation =2=2cos36 + cos 72 .

(e) Use the Pythagoras theorem to deduce that 1 √ 1 √ sin 72◦ = 10 + 2 5; sin 36◦ = 10 − 2 5. 4 4 YIU : Trigonometry Workbook 49

θ 18◦ 36◦ 54◦ 72◦ √ √ √ √ Summary sin θ 1 ( 5 − 1) 1 10 − 2 5 1 ( 5+1) 1 10 + 2 5 4 √ 4 √ 4 √ 4 √ θ 1 1 1 − 1 − cos 4 10 + 2 5 4 ( 5+1) 4 10 2 5 4 ( 5 1)

Hints and Comments ◦ ◦ 1(a) ABP is isosceles with vertical angle 30 .Itfollowsthat√ ABP√ =75,and PBM = ◦ ◦ ABP − 60 =15 .√ By Pythagoras√ Theorem, AD = 3sothatDE =2− 3. ◦ 1 (b) cos 15 = 4 ( 6+ 2). √ (c) 75◦ is the complement of 15◦. tan 75◦ =2+ 3. 2(c). The cosine function is decreasing, i.e., greater acute angles have smaller cosine functions. The lesson of these examples is that the trigonometric functions are not additive: you cannot write T (θ + ϕ)=T (θ)+T (ϕ) for any of these functions. YIU : Trigonometry Workbook 50

4.3 Subsidiary functions

For an acute angle θ, we deﬁne the contangent, secant,andcosecant functions by

1 cos θ cot θ := = ; tan θ sin θ 1 sec θ := ; cos θ 1 csc θ := . sin θ These subsidiary functions are helpful in simplifying expressions involving the basic sine and cosine functions. It is important to realize that any one of the 6 trigonometric functions of an acute angle determines the remaining 5. 1. (Alternative versions of Pythagoras Theorem)

tan2 θ +1 = sec2 θ; 1+cot2 θ =csc2 θ.

θ sec θ 2. Show that tan = csc θ . YIU : Trigonometry Workbook 51

3. Complete the following table (without ﬁnding the measure of the angle θ).

sin θ cos θ tan θ cot θ sec θ csc θ 12 (i) 37 16 (ii) 65 55 (iii) 48 80 (iv) 39 41 (v) 9 85 (vi) 36

4. Solve the following right triangles with C =90◦,givenA and one side indicated by an asterisk. (Of course, B =90◦ − A).

abc (i) * (ii) * (iii) *

4. In each of the following cases, ﬁnd an acute angle with given trigonometric function.

Given θ (i) sin θ =0.4 (ii) cos θ =0.6 (i) tan θ =0.5 (i) cot θ =0.4 (i) sec θ =1.4 (i) csc θ =1.6

5. Why is it not possible to ﬁnd an acute angle whose sine function is 1.4 ? How about one with secant function 0.7 ? YIU : Trigonometry Workbook 52

6. For each number r between 0 and 1. Explain why it is always possible to ﬁnd an acute angle θ such that sin θ = r. Why is this angle unique ? What if sine is replaced by cosine ?

7. Let r be any real number. Explain why it is always possible to ﬁnd an acute angle θ such that tan θ = r. Why is this angle unique ? What if tangent is replaced by cotangent ?

8. Let r ≥ 1. Explain why it is always possible to ﬁnd an acute angle θ such that sec θ = r. Why is this angle unique ? What if secant is replaced by cosecant ? YIU : Trigonometry Workbook 53

Hints and Comments 1. Start with the relation sin2 θ+cos2 θ = 1. Divide both sides by cos2 θ to arrive at tan2 θ+1 = sec2 θ. 3. In each case, realize the given trigonometric function as the ratio of two sides of a right triangle. Determine the third side by Pythagoras Theorem, and write down the other trigonometric functions. The answers to Exercise 1.2.2 should be helpful. 4. (i) b = a cot A, c = a csc A; (ii) a = b tan A, c = b sec A; (iii) a = c sin A, b = c cos A. Chapter 5

Simple Problems in 2 and 3 Dimensions

5.1 Heights and distances

1. Let P be a point on a semicircle with diameter AB such that PAB = θ. What is the area of AP B ?

2. The centers of two circles of radii 3 cm. and 5 cm. are 12 cm. apart. (a) Find the angle between the two external common tangents.

(b) How about the angle between their internal common tangents.

54 YIU : Trigonometry Workbook 55

3. Two circles of radii 3 cm. and 5 cm. are touching each other externally. (a) Find the angle between the two external common tangents.

(b) What is the distance between the points of contact of the circles with a common tangent ?

4. A sphere of radius 8 cm. is put inside a right circular cone whose axis is vertical. The highest point of the sphere is 22 cm. above the vertex of the cone. Find the (semi- vertical) angle of the cone.

5. Identical isosceles triangles are cut oﬀ from the four corner of a square to form a regular octagon. How big is each of these triangles ? YIU : Trigonometry Workbook 56

6. The upper part of a tree broken over by the wind makes an angle 30◦ with the ground and its top rests on the ground at a point 100 feet from the root. What was the height of the tree ?

7. At the top of a house 100 feet high, the angle of elevation of the top of a tower is 45◦; on the ground ﬂoor it is 60◦. Find the height of the tower.

8. At a point half way between two buildings the angles of elevation of their tops are 30◦ and 60◦. Show that one building is three times as tall as the other.

9. A conical funnel, vertical angle 50◦, rests inside a glass of height 7 inches and diameter 4 inches, internal measurements. Find the height of the apex of the funnel above the base of the glass. YIU : Trigonometry Workbook 57

10. The vertical angle of an isosceles triangle is 2θ, and its circumradius is R.What is the area of the triangle ?

11. Suppose the inradius is r. What is the area of the triangle ?

12*. Let C be the midpoint of a semicircle with diameter AB, center O and radius r.LetP be a point between O and B,andθ := OCP. Through P construct the perpendicular to AB meeting BC at Q. Show that (i) AQP = θ; (ii) AP = a(1 + tan θ); (iii) PQ= a(1 − tan θ). ◦ θ 1+tan θ Deduce that tan(45 + )= 1−tan θ .

13*. The centers of n identical circles each of radius are at the vertices of a regular n−gon each of side 2r (so that each circle is tangent to its two immediate neighbours). Find the area enclosed by the circles. YIU : Trigonometry Workbook 58

Hints and Comments 1 2 1. 2 AB sin θ cos θ. 5. Suppose each side of the square has unit length, and each (shorter)√ side of the isosceles triangles√ is x ◦ x x 1√ 1 − . Since each external angle of a regular octogon is 45 ,wehave(2+ 2) =1; = 2+ 2 = 2 (2 2) = 0.293 ···. √ ◦ 1 6. 100(tan 30 + cos 30◦ )=···= 100 3. 7. Let A be a viewer and B an object. According as B is above or below A, the angle of elevation (respectively depression)fromA to B is the angle between the horizontal through A and the line AB. Suppose the distance from the tower to√ the house is x ft. and the height√ of the tower is y ft. Then x(tan 60◦ − tan 45◦) = 100; x = √100 = 50( 3+1);y = xtan60◦ = 50(3 + 3). √ √3−1 ◦ ◦ 1 8. tan 60 : tan 30 = 3: 3 3=3:1. 9. 7 − 2 cot 25◦. 10. 4R2 sin θ cos3 θ. (1+sin θ)2 2 11. 2sinθ cos θ r . 2π 13*. The regular n−gon consists of n isosceles triangle each of vertical angle n and base 2r.Ithas 1 r nr2 (n−2)π area n · · 2r · π = π . At each vertex there is a circular sector of radius r and angle .These 2 tan n tan n n 1 2 (n−2)π (n−2)π 2 n circular sectors together have area n· 2 r n = 2 r . The area enclosed by the chain of n circles is therefore n (n − 2)π 2 2 π π 2 π − r . = πr + n(cot − )r tan n 2 n 2 π π Note that this area is greater than the circle of radius r if cot n > 2 .Thisistrueifn ≥ 6.

n √345678910√ π 1 cot n 3 3 1 1.3764 3 Area r = 1 0.161254 0.858407 2.16952 4.10912 6.68167 9.88893 13.7317 18.2105 YIU : Trigonometry Workbook 59

5.2 Problem in 3 dimensions

Angle between a line and a plane Suppose a straight line l and a plane P meet at apointA. To ﬁnd the angle between the line and the plane, choose any point P on the line l, and draw a perpendicular from P to the plane, meeting the latter at a point Q. Then PAQ is the angle between l and P.

Angle between two planes Suppose two planes P1 and P2 intersect in a line l.IfA is a point on l and AX, AY are lines in the planes P1, P2 respectively perpendicular to l, then XAY is the angle between the two planes.

Exercises 1(a). The angle of elevation of a ballon from a station P due south of it is 60◦,and that from another station Q due west (of the ballon) is 45◦. The two stations are 2 miles apart. Find the height of the ballon.

1(b). The angle of elevation of a ballon from a station P due south of it is 60◦,and from another station Q due west from P and 2 miles from it the angle of elevation is 45◦. YIU : Trigonometry Workbook 60

Find the height of the ballon.

2. Let ABCDEF GH be a cube, with AE, BF, CG, DH perpendicular to the square ABCD. (a) Calculate the angle between the two diagonals AG and BH.

(b) What is the angle between the diagonal AG and the base ABCD ?

3. A rectangle box has dimensions 8 cm. × 9cm. × 12 cm. Find the angle between a diagonal and each of the 9 cm. × 12 cm. face.

4(a). Each side of a regular tetrahedron has length a. YIU : Trigonometry Workbook 61

What is the height of the solid ?

4(b). Three identical billiard balls of radii r are touching one another on a horizontal table. A fourth identical, billiard ball is placed on top of these three. Find the height of the center of this fourth billiard ball above the table.

5. Find the angle between an edge and a face of a regular tetrahedron.

6. Find the angle between two faces of a regular tetrahedron.

7. A right circular cone is formed by glueing together the two outermost radii of a 120◦ sector. The radii of the sector become the generators of the cone. Find the angle YIU : Trigonometry Workbook 62 between a generator and the base circle.

8. A right pyramid has a square base of side 12 cm, and height 7 cm. Find the angle between a slant edge and the square base.

9. A right pyramid has a square base and other 4 faces equilateral triangles. (a) Find the angle between a slant edge and the square base.

(b) Find the angle between a slant face and the square base.

10. A book 8 in. × 10 in. rests on a horizontal table. Its frontcover is opened to make an angle 120◦ with the ﬁrst page. What is the angle between a diagonal of this frontcover with the horizontal ?

11. An isosceles triangle with vertical 30◦ makes an angle 60◦ with the horizontal. What is the angle between a slant side of the triangle with the horizontal ?

12. An equilateral triangle is in a vertical position, with its base horizontal. Under the sun its shadow is an isosceles triangle of vertical angle 30◦. What is the angle of elevation of the sun ?

13. The projection of an isosceles triangle of semi-vertical angle α on a horizontal plane is an isosceles triangle of semi-vertical angle β. (a) What is the angle between the isosceles triangle and the horizontal plane ? YIU : Trigonometry Workbook 64

(b) What is the angle between a slant side of the isosceles triangle with the horizontal plane ?

14. The base of a right pyramid is a regular pentgon, and its slant faces are equilateral triangles. (a) What is the angle between a face of the pyramid and the base ?

(b) What is the angle between a slant side of the pyramid and the base ?

15*. The base of a right pyramid is a regular n−gon, and its slant faces are isosceles YIU : Trigonometry Workbook 65

90 ◦ triangles of semi-vertical angles n . (a) What is the angle between a face of the pyramid and the base ?

(b) What is the angle between a slant side of the pyramid and the base ?

Hints and Comments 1. Let x be the height of the ballon and a the distance between the two stations. If α and β are 2 2 2 2 a2 the angles of elevation at P and Q respectively, (a) (x cot α) +(x cot β) = a , x = 2 2 ;(b) cot α+cot√β x α 2 a2 x β 2 x2 a2 a α ◦ β ◦ x ( cot√ ) + =( cot ) ; = cot2 α−cot2 β .If =2, =60 and =45,then(a) = 3, (b) x = 6. AG BH P 2. Each of the diagonals√ and bisect each other, say, at . If each side of the cube has length 3 a,thenAP = BP = 2 a. (a) The angle between the diagonals AG and BH is AP B.InAP B, AP B =2arcsin AB =2arcsin√1 = ···=70.53◦.(b)arcsinCG =arcsin√1 =35.27◦. 2AP 3 AG 3 √ 3. Consider a cuboid of dimensions a × b × c. The length of a diagonal is a2 + b2 + c2.The b × c √ a a b c angle between a diagonal and a face is arcsin a2+b2+c2 .With =8, =9, = 12, this is arcsin 817 = 28.08◦. YIU : Trigonometry Workbook 66

4(a). Let ABCD be a regular tetrahedron each side of length a.IfO is the center of ABC,then √ √ √a 2 2 2 6 OA = . The height of the tetrahedron is OD = a − OA = 3 a = 3 a ≈ 0.816a. 3 √ √ 6 3+2 6 4(b). r + 3 · 2r = 3 r ≈ 2.633a. OAD √1 . OAD . ◦ 5. cos = 3 =05774; =5474 . P AB CP DP AB CP 6. Let√ be the midpoint of .Then and are each perpendicular to .Notethat = DP 3 a ABC ABD CPD 1 CPD √1 . = 2 . The angle between the faces and is ,andsin2 = 3 =05774. It follows that CPD =70.53◦. 120 1 7. Let r be the radius of the sector. The base radius of the cone is 360 r = 3 r.Ifθ is the angle 1 ◦ between a generator and the base, cos θ = 3 . θ =70.53 . √7 . . ◦ 8. arctan 6 2 =arctan08250 = 39 52 . 9. Let ABCD be the square base and P the vertex of the pyramid. AP = AB = ···= a.LetO be the center of the√ square base. √ (a) OA = 2 a.cosOAP = 2 ; OAP =45◦. 2 2 √ √ 2 (b) It follows that the height of the pyramid is 2 .IfM is the midpoint of AB,thentanOMP = 2; OMP =54.74◦. Q AP BQ CQ AP (c) Let √be the midpoint√ of .Then√ and are each perpendicular to .Notethat BQ CQ 3 BD 1 BQD √2 . BQD . ◦ = = 2 and = 2. sin 2 = 3 =08165; = 109 47 . 10. Suppose we have a book a in. × b in., bound along a side of b in. If the front cover is open to ◦ make an angle θ (or 180 − θ) with the front√ page, the height of the opposite of the binding edge is a sin θ. The angle Since the length of a diagonal is a2 + b2, the angle between the diagonal and the horizontal is a sin θ arcsin √ . a2 + b2 If a =8,b = 10, and θ =60◦,thisis32.76◦. 11. Let h be the height of an isosceles triangle of semi-vertical angle θ. If the triangle makes an angle h ϕ with the horizontal, the vertex is h sin ϕ above the horizontal. The length of a slant side being cos θ ,it makes an angle ω with the horizontal, where h ϕ ω sin ϕ θ. sin = h =sin cos cos θ with θ =15◦ and ϕ =60◦,thisis56.78◦. 12. More generally, consider an isosceles triangle of semi- vertical angle α with its base horizontal. Suppose the length of the base is 2b, and that the shadow of the triangle is an isosceles angle of semi- verticalangle β. Then, the height of the triangle is a cot α and its shadow has length a cot β. The angle of elevation ω is given by a cot α tan β tan ω = = . a cot β tan α With α =30◦ and β =15◦,thisisω =24.90◦. 13. Suppose the base of the isosceles triangle is 2a. (a) The heights of the triangle and its projection are respectively a cot α and a cot β. The angle between the two planes is the angle between these two a cot β tan α heights, and is arctan a cot α =arctantan β . (b) The lengths of slant sides of the isosceles triangle and its projection are respectively a csc α and a csc β. The angle between a slant side of the triangle and the a csc β sin α horizontalis arccos a csc α = arccos sin β . YIU : Trigonometry Workbook 67

◦ ◦ tan 30◦ 14. In the notation of Exercise 13, put α =30 and β =36. (a) arctan tan 36◦ =arctan0.7947 = ◦ sin 30◦ ◦ 38.47 . (b) arccos sin 36◦ = arccos 0.8507 = 31.72 .(c)LetP be the vertex, and A, B, C three consecutive vertices of the regular pentagon. Note that the projections of PA and PC on the edge PB coincide, say, Q AQC PAB PBC a at , is the angle between the faces √ and .Let be the length of a side of the regular ◦ 3 ◦ pentagon. Then, AQ = CQ = a sin 60 = 2 a. On the other hand, AC =2a sin 54 . In the isosceles ◦ 1 AC sin 54 ◦ triangle QAC,sin2 AQC = 2AQ = sin 60◦ =0.9342. From this, AQC = 138.19 . 90 ◦ 15. Let α := n . The projection of a slant face on the base is an isosceles triangle of semi-vertical ◦ tan α sin α angle 2α . (a) The angle between a face and the horizontal is αn := arctan tan 2α .(b)βn := arccos sin 2α . (c) In the notation of Exercise 14(c), the angle between two adjacent faces of the pyramid is γn := cos 2α 2arcsin cos α .

nan αn =arctanan bb βn = arccos bn cn γn =2arcsincn 33 71.5651◦ 0.57735 54.7356◦ 0.57735 70.5288◦ 4 2.41421 67.5◦ 0.541196 57.2349◦ 0.765367 99.8793◦ 5 2.23607 65.9052◦ 0.525731 58.2825◦ 0.850651 116.565◦ 6 2.1547 65.1039◦ 0.517638 58.826◦ 0.896575 127.423◦ 7 2.10992 64.6413◦ 0.512858 59.1456◦ 0.924139 135.078◦ 8 2.08239 64.3489◦ 0.509796 59.3498◦ 0.941979 140.773◦ 9 2.06418 64.1519◦ 0.507713 59.4884◦ 0.954189 145.18◦ 10 2.05146 64.0127◦ 0.506233 59.5868◦ 0.962912 148.693◦

◦ ◦ Indeed, if n is large, αn is close to arctan 2 = 63.57 , βn is close to 60 . What is the corresponding limit for γn ? Chapter 6

Volumes of Solids

6.1 Parallelepiped and tetrahedra

Basic volume principles (A) The volume of a cuboid (a parallelepiped whose faces are rectangles) of dimensions a, b and c units is abc cubic units. (B) Suppose two solids are intercepted between the same two horizontal planes. If at each horizontal level, the areas of the sections of the two solids are equal, then the volumes of the solids are equal. (C) Volume of parallelepiped = Area of base parallelogram × height.

68 YIU : Trigonometry Workbook 69

Exercises 1. Consider a pyramid with base a parallelogram XY ZW and vertex P as made up of two tetrahedra T1 := XY ZP and T2 := XWZP. (a) What is the intersection of the pyramid with a plane parallel to the base ?

(b) What are the intersections of the same plane with the tetrahedra T1 and T2 ?

2. Consider a tetrahedron ABCD. (i) In the plane of the base ABC, complete the parallelogram ABCE. (ii) In the plane of the slant face BCD, complete the parallelogram DBCF.

(a) Why is DAEF a parallelogram ? YIU : Trigonometry Workbook 70

(b) What is the volume of the prism ABDECF in terms of the area of ABC and the height of the tetrahedron from D to the base ?

(c) For convenience, we adopt the notation (ABC, D)forthevolume of the tetrahedron with base triangle ABC and vertex D. Explain each of the following equalities. (i) (ABC, D)=(ACE, D); (ii) (ACE, D)=(ADE, C); (iii) (ADE, C)=(DEF, C). (d) Make use of the fact that the prism ABDECF is made up of the three tetrahedra ABCD, ACED, CDEF to conclude that the volume of the tetrahedron ABCD is equal 1 × ABC× D to 3 area of base height (from to the base).

1 × × Volume of tetrahedron = 3 area of base triangle height. YIU : Trigonometry Workbook 71

3. Each side of a regular tetrahedron has length a. Find the volume of the tetrahedron.

Hints and Comments 1. (a) parallelogram; (b) congruent triangles making up the parallelogram in (a); (c) at each level, the sections of the two tetrahedra have equalareas. 2(a) AE and DF are equal and parallel. 1 2(b) Volume of prism = 2 × area of parallelogram ABCE × height = ABC× height. 2(c) (i) and (iii) follow from Exercise 1(c). (ii): same tetrahedron in diﬀerent notations. 2(d) Volume of prism = (ABC, D)+(ACE, D)+(DEF, C)=3(ABC, D) by (c) above. It follows ABC, D 1 ×ABC× D that ( )= 3 height from to this√ base. 3 2 3. Its base is an equilateral triangle with area 4 a . The projection of a vertex on its opposite face √ √a 2 a 6 a is the center of this face. Since the radius of the base is 3 , the height of the tetrahedron is 3 = 3 . √ √ 1 3 2 2 2 3 The volume is 3 · 4 a · 3 = 12 a . YIU : Trigonometry Workbook 72

6.2 Pyramids

A pyramid is a solid formed by joining points in and on a polygonal base to a vertex outside the plane of the polygon.

1 × × Volume of pyramid = 3 base area height.

Exercises 1. The base of a pyramid is a square each side of length a. Its slant faces are equilateral triangles. What is the volume of the solid ?

2. The base of a pyramid is a regular hexagon of side 6 cm. and its height is 8 cm. Find the angle between (i) a slant edge with the base, (ii) a slant face with the base. YIU : Trigonometry Workbook 73

3. The base of a pyramid is a regular n−gon inscribed in a circle radius a,andits α<2π slant faces are all isoceles triangles of vertical angles n . Find the angle between (i) a slant edge with the base, (ii) a slant face with the base.

Hints and Comments ABCD O P O 1. Let be the square base of√ the pyramid and the outside vertex, the projection√ of OA a AP 2 a 2 a on the base. Note that = , = 2 √ . It follows√ that the height of the pyramid is also 2 .From 1 · a2 · 2 a 2 a3 this the volume of the pyramid is 3 2 = 6 . √ 2. The circumcircle of the hexagon has radius 6 cm. Each slant edge has length 62 +82 =10cm. 6 . ◦ It follows that the angle between each slant edge√ and the base√ is arccos 10 =3687 .Now,itiseasyto see that the height of each slant face has length 102 − 32 = 91 cm. The angle between each slant face √8 and the base is arcsin 91 =. π 3. Each side of the polygon has length 2a sin n , and the distance from the center to the midpoint of a π π α π α side is a cos n . The height of each slant face is a sin n cot 2 , and each slant edge has length a sin n csc 2 . The angle between a slant edge and the base is α sin 2 arccos π . sin n

Theanglebetweenaslantfaceandthebaseis π α a cos n tan 2 arccos π α = arccos π . a sin n cot 2 tan n YIU : Trigonometry Workbook 74

6.3 Right circular cones

Let r bethebaseradius,h the height, and l the length of a slant edge of a circular cone. These lengths are related by l2 = r2 + h2. By cutting the cone along a generator and ﬂattening it on a plane, we ﬁnd that the

Surface area of right circular cone = πrl; 1 πr2h Volume of right circular cone = 3 .

Exercises 1. A circular sector of radius 10 cm. and angle 216◦ is bent into a right circular cone. Find the volume of the cone.

2. A right circular cone has base radius 6 cm and height 8 cm. An ant is crawling on the conical surface from a point on the base to its antipode (diametrically opposite) point. What is the shortest distance ? YIU : Trigonometry Workbook 75

3. Find the volume of a right circular cone of semi-vertical angle θ and base radius r.

4. A circular sector of angle α is bent into a circular cone. What is the semi-vertical angle of the cone ?

5(a). Find the radius of the largest sphere contained in a right circular cone of height h and semi-vertical angle α. What is this in terms of α and the radius R of the base ?

5(b). Find the base radius of the largest right circular cone of semi-vertical angle α that can be contained in a sphere of radius R. YIU : Trigonometry Workbook 76

6. A right circular cone is cut by two planes parallel to its base through the points of trisection of its height. Find the ratio of (i) areas of the curved surfaces and (ii) volumes of the two resulting frusta.

7. An inverted conical container of semi-vertical angle α contains water up to level d1. How much more water should be poured in to make the water level rise to d2 ?

Hints and Comments 216 · π · π 1. The circumference of the√ base circle of the cone being 360 2 10 = 12 cm., the radius of the 2 2 π 2 3 base is 6 cm, and the height = 10 − 6 = 8 cm. The volume of the cone is 3 · 6 · 8=96π cm . 2. Note that this is not the semicircumference of the base circle. Instead, ﬂatten the cone and represent√ such a shortest path on the conicalsurface by a straight line segment. The radius of the sector being 62 +82 = 10 cm, the angle of the sector from which one makes the cone is 1 3 3. 3 r cot θ. 4. If l is the radius of the sector, and r the radius of the base circle, then 2πr = αl.Fromthis, r α α l = 2π . The semi-verticalangleof the cone is arcsin 2π . 1 sin α 5(a). If the radius of the sphere is r,thenr(1 + csc α)=h; r = 1+csc α h = 1+sin α h.Intermsofthe cot α cos α π α base radius, this is r = 1+csc α R = 1+sin α R. (This can also be written as R tan( 4 − 2 )). 5(b). r =2R sin 2α. 6(a). Flatten the cone into a circular sectors, and see that the areas of the curved surfaces are in the ratio 32 − 21 :22 − 12 =5:3. 6(b). 33 − 23 :23 :13 =19:7. 1 2 1 2 1 3 3 2 7. 3 (d2 tan α) d2 − 3 (d1 tan α) d1 = 3 (d2 − d1)tan α. YIU : Trigonometry Workbook 77

6.4 Spheres

The volume of a sphere can be determined by using the basic volume principle. We do this by considering a hemisphere of radius r, as a solid between two horizontal planes, one containing the equator and the other the tangent plane through the north pole. Between the same two planes, consider a cylinder of the same radius r.Fromthe interior of the cylinder remove an inverted right circular cone whose base circle is the section of the cylinder at the level of the north pole of the hemisphere, and whose vertex is the center of the section at the level of the equator. This right circular cone of course has semi-vertical angle 45◦, so that at each horizontal level, the radius of the section is the same as the height above the equatorial plane.

Consider the horizontal sections of the two solids at a height x above the equatorial plane. From the hemisphere, this is a circle, whose radius a := a(x)isgivenbya2+x2 = r2. The area of this section of the hemisphere is therefore πa2 = π(r2−x2). The corresponding section of the other solid is πr2 − πx2 = π(r2 − x2). These sections have the same area. By the basic volume principle,

Volume between equator and level x of hemisphere = volume between cylinder and cone of height x 1 = πr2 · x − πx2 · x π 3 = x(3r2 − x2). 3 x r 2π r3 In particular, with = , we obtain the volume of the hemisphere = 3 . This leads to the formula YIU : Trigonometry Workbook 78

4π r3 r Volume of sphere = 3 , =radius.

Exercises 1. A spherical bowl of radius r contains water up to level x ≤ 2r. Find the amount of water.

2. A solid consists of a right circular cone mounted on a hemi-spherical base. Find the semi-vertical angle of the cone if the volumes of the conical and the spherical portions are equal.

3. Find the ratio of the volumes of a right circular cone of semi-vertical angle 30◦ and the largest sphere that can be put into it.

4. A metal ball of radius r sinks into conical tank of semivertical angle α, containing water up to a depth h. By how much does the water level rise ? YIU : Trigonometry Workbook 79

5. A metal ball of radius r just submerges under water in an inverted conical tank. What will be the depth of water in the cone if the ball is removed ?

6. Four identical spheres of radii r are touching each other. Find the volumne enclosed between them.

Hints and Comments 4 3 π 2 2 π 2 1. 3 πr − 3 (r − x)(3r − (r − x) )= 3 (3r − x)x . 2 3 2. Let r be the radius of the hemisphere. The volume of the hemisphere = 3 πr = volume of the 1 ◦ cone. It follows that the height of the cone is 2r. The semiverticalangleof the cone is arctan 2 =26.57 . 3. Let α be the semi-verticalangleand h the height of the cone. The radius of the sphere is sin α r = 1+sin α h. The volumes of the cone and the sphere are in the ratio

1 4 sin3 α sin3 α πh3 tan2 α : π =tan2 α :4 3 3 (1 + sin α)3 (1 + α)3 =(1+sinα)3 :4sinα cos2 α =(1+s)3 :4s(1 − s2) =(1+s)2 :4s(1 − s)

◦ 1 3 2 where s := sin α.Ifα =30 , s = 2 , these are in the ratio ( 2 ) :1=9:4. d 1 πd3 2 α 1 πh3 2 α 4 πr3 d3 − h3 r3 2 α 4. Let be the new depth of water.√ 3 tan = 3 tan + 3 ; =4 cot .The 3 rise in water level = d − h = h + r 4cot2 α. 5. Let s := sin α, α = semi-vertical angle of the cone. Volume of cone : volume of ball = (1 + s)2 : 2 1−2s+5s2 4π 3 4s(1−s). Volume of water : volume of ball = (1−2s+5s ):4s(1−s). Amount of water = 4s(1−s) · 3 r . YIU : Trigonometry Workbook 80

If the depth of water is d,then

π 1 − 2s +5s2 4π d3 tan2 α = · r3; 3 4s(1 − s) 3 (1 − 2s +5s2) 1 − s2 (1 + s)(1 − 2s +5s2) d3 = · r3 = r3. s(1 − s) s2 s3 √ ◦ 1 3 3 3 If α =30 , s = 2 , d =15r ,andd = 15r. Chapter 7

Trigonometric functions of general angles

7.1 Basic properties

Let P (x, y) be a point in a rectangular coordinate system. If P does not coincide with the origin O,wedenotebyρ the length of OP,whichisthepositive square root of x2 + y2. The vector OP has direction speciﬁed by the directed angle AOP ,whereA is the point with coordinates (1,0) on the x−axis, and the angle is positive or negative according as the rotation from OA to OP is counter – clockwise or clockwise. x y θ θ . sin := ρ;cos:= ρ

81 YIU : Trigonometry Workbook 82

Pythagoras Theorem For any angle θ,

sin2 θ +cos2 θ =1.

The sine and cosine functions together completely determine an angle within 2π (radians). In this sense they are the primary trigonometric functions, and the tangent function can be regarded as secondary: sin θ tan θ := . cos θ We shall also consider the three tertiary trigonometric functions 1 cos θ 1 1 cot θ := = ;secθ := ;cscθ := . tan θ sin θ cos θ sin θ

Exercise 1. Complete the following tables of values of trigonometric functions.

θ degrees sin θ cos θ tan θ π ◦ (i) 2 90 2π ◦ (ii) 3 120 (iii) 180◦ (iv) 225◦ (v) 270◦ (vi) 300◦ (vii) 2π

2. The minute–hand of a clock, whose face is in a vertical plane, is 6 inches long. Find the distance of the tip from the central vertical line of the clock at t minutes past the hour.

3. (Sign distribution of the trigonometric functions in the four quadrants) YIU : Trigonometry Workbook 83

Quadrant x−coordinate y−coordinate sin θ cos θ tan θ First I + + + + + (a) Second II − + Third III −− Fourth IV + −

(b) The sine function is positive in the first and quadrants and negative in the and quadrants. The cosine function is positive in the first and quadrants and negative in the and quadrants. The tangent function is positive in the first and quadrants and negative in the and quadrants. 4. (Calculation of trigonometric functions in terms of acute angles) (a) Every angle in the second quadrant can be written in the form π − θ for an acute angle θ. Draw a diagram to explain the following.

sin(π − θ)=sinθ; cos(π − θ)=− cos θ; tan(π − θ)=− tan θ.

(b) Every angle in the third quadrant can be written in the form π + θ for an acute angle θ. Draw a diagram to explain the following.

sin(π + θ)=− sin θ; cos(π + θ)=− cos θ; tan(π + θ)=tanθ.

(c) Every angle in the fourth quadrant can be written in the form 2π − θ for an acute angle θ. Draw a diagram to explain the following.

sin(2π − θ)=− sin θ; cos(2π − θ)=cosθ; tan(2π − θ)=− tan θ.

5. Let θ be an acute angle. With the help of a simple diagram, determine the trigono- π θ θ metric functions of 2 + etc in terms of those of . YIU : Trigonometry Workbook 84

sin cos π − θ θ θ 2 cos sin π θ 2 + 3π − θ 2 3π θ 2 +

6. (Periodicity)

sin(θ +2π)=sinθ; cos(θ +2π)=cosθ; tan(θ + π)=tanθ.

7. (Even and odd properties)

sin(−θ)=− sin θ; cos(−θ)=cosθ; tan(−θ)=− tan θ. YIU : Trigonometry Workbook 85

7.2 Inverse trigonometric functions

Let a ∈ [−1, 1], i.e., −1 ≤ a ≤ 1. There is a unique angle α a − π ≤ α ≤ π := arcsin in the range 2 2 satisfying sin α = a. This is the principal value of the inverse sine function,andis also denoted by sin−1 a.Withthis,every solution of the equation sin θ = a is of the form

θ = kπ +(−1)k arcsin a, k ∈ Z.

See the graph on p.88(a). Similarly, for every a ∈ [−1, 1], there is a unique angle

β := arccos a in the range 0 ≤ α ≤ π satisfying cos β = a. This is the principal value of the inverse cosine function,andis also denoted by cos−1 a. 1. Every solution of cos θ = a is of the form

θ =2kπ ± arccos a, k ∈ Z.

See the graph on p.88(b). Finally, for every real number a,thereisaunique angle γ a − π <α< π := arctan in the range 2 2 satisfying tan γ = a. Thisistheprincipal value of the inverse tangent function,and is also denoted by tan−1 a. 2. Every solution of tan θ = a is of the form

θ = kπ + arctan a, k ∈ Z.

See the graph on p.88(c). YIU : Trigonometry Workbook 86

3. Complete the following table.

a arcsin arccos arctan 1 (i) √2 2 (ii) 2√ − 3 (iii) 2 (iv) 1 (v) 2 (vi) -3

4(a). arcsin(−a)=− arcsin a. 4(b). arccos(−a)= . 4(c). arctan(−a)= . 5. Suppose α =arcsina for a ∈ [−1, 1]. What is arccos a in terms of α ? YIU : Trigonometry Workbook 87

7.3 The sine formula

Consider the circumcircle of ABC,withcircumcenter O and circumradius R.Let D be the midpoint of the side BC.

A BOD A R a 1(a). Assume acute. Show that = and 2 = sin A .

A BOD R a 1(b) If is obtuse, what is ? Show that 2 = sin A still holds. YIU : Trigonometry Workbook 88

From this, we deduce the sine formula: for any triangle ABC, a b c = = (= 2R). sin A sin B sin C

Example (Solution of triangles: AAS case) Suppose A, B and a are given. Then, clearly, C = π − (A + B), and from the sine formula, a sin B a sin C a sin(A + B) b = ; c = = . sin A sin A sin A 2. Solve the following triangles.

AB a Cb c (i) 50◦ 60◦ 10 (ii) 50◦ 50◦ 10 (iii) 50◦ 40◦ 10 (iv) 50◦ 70◦ 10 (v) 50◦ 80◦ 10

a b A B b sin A < B ≤ Example Suppose , and are given. Then sin = a .Since0 sin 1, this requires b sin A ≤ a for the existence of such a triangle. b A a B B π (i) If sin = ,thensin =1,and = 2 . In this case, we have a right triangle. What is c in terms of a, b and A in this case ?

b Aa β>A Note that = also satisﬁes sin = a .If ,then and there is B π −β C π − A π −β β −A c a sin C a sin(β−A) another triangle with = , = ( + )= ,and = sin A = sin A . YIU : Trigonometry Workbook 89

Summary The number of triangles ABC with given a, b,andA is (i) none if a

abABC c (i) 4 6 30◦ (ii) 3 6 30◦ (iii) 2.5 6 30◦ 4. Find the circumradius of each of the (iv) 4 5 60◦ (v) 4 4.5 60◦ (vi) 4 3.5 60◦ triangles in Exercise 3.

(i) (ii) (iii) (iv) (v) (vi) circumradius R

1 ab C 5(a) Show that the area of a triangle is given by = 2 sin .

R abc 5(b). Show that the circumradius of a triangle is given by = 4 . YIU : Trigonometry Workbook 90

6. Find the circumradius of the following triangles.

abc R (i) 3 4 5 (ii) 4 5 6 (iii) 13 14 15 (iv) 5 12 13 (v) 17 18 19

Hints and Comments 3. (i) B =48.59◦, c =7.842, C = 101.4◦,orB = 131.4◦, c =2.55, C =18.59. (ii) B =90◦. (iii), (iv) no solution. (v) B =76.98◦, c =3.151, C =43.02◦ or B = 103◦, c =1.349, C =16.98◦. (vi) Only one solution, B =49.27◦, c =4.36, C =70.73◦. 4. (i) R = 4 for both triangles; (ii) R =3;(v)R =2.31 for both triangles; (vi) R =2.31. 1 c 5(b). Note that the area of ABC canbewrittenas = 2 ab sin C.SinceR = 2sinC , by eliminating abc C,wehaveR = 4 . 6. abc R (i) 3 4 5 6√ 2.5√ 15 8 (ii) 4 5 6 4 7 7 7 65 (iii) 13 14 15 84 8 (iv) 5 12 13 30√ 6.5√ 323 (v) 17 18 19 36 15 120 15 YIU : Trigonometry Workbook 91

7.4 The cosine formula

Solution of triangles: SAS case.Givena, b and C, we determine c by using the cosine formula, and then A and B by using the sine formula.

The most natural solution of this problem is to set up a coordinate system in which we put the vertex C at the origin O, the vertex A along the x−axis, with coordinates (b, 0), the vertex along the direction given by angle C,withOB = CB = a.Notethat the coordinates of B are (a cos C, c sin C). The side c of ABC is the distance between A(b, 0) and B,andisgivenby

c2 =(a cos C − b)2 +(a sin C)2 = a2(cos2 C +sin2 C) − 2ab cos C + b2 = a2 + b2 − 2ab cos C.

Thecosineformula: c2 = a2 + b2 − 2ab cos C. Equivalently,

a2 + b2 − c2 cos C = . 2ab YIU : Trigonometry Workbook 92

Exercises ab C cA B (i) 13 14 60◦ (ii) 21 41 50◦ 1. Solve the following triangles. (iii) 39 41 120◦ (iv) 25 51 40◦ (v) 25 63 45◦

Example (Solution of triangles: SSS case) Given the three sides a, b,andc of a triangle, the three angles are determined by

b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 cos A = ;cosB = ;cosC = . 2bc 2ca 2ab 2. Solve the following triangles.

ab c AB C (i) 13 14 15 (ii) 21 41 50 (iii) 39 41 50 (iv) 25 51 74 (v) 25 63 74

3. The centers of two circles, radii 17 cm. and 25 cm., are 28 cm. apart. Find the angle between the tangents at an intersection of the two circles.

4. The hour–hand and the minute–hand of a clock are respectively 3 inches and 4 inches long. (a) Find the distance between the ends of the two hands at t minutes past k O’Clock. YIU : Trigonometry Workbook 93

(b) At what times are these two ends exactly 5 inches apart ?

Hour123456789101112 Minute Minute

Example (Alternative solution of the SSS case) Given the lengths a, b, c of the three sides of a triangle ABC, show that A (s − b)(s − c) B (s − c)(s − a) C (s − a)(s − b) tan = , tan = , tan = . 2 s(s − a) 2 s(s − b) 2 s(s − a)

The exradii of the triangle are given by r ,r ,r . A = s − a B = s − b C = s − c 5. Three circles of radii 1, 2, 3 inches are mutually tangent to each other externally. Find the area between the three circles.

1 ab C Example* (Alternative proof of the Heron formula) Since = 2 sin ,to ﬁnd the area in terms of the sides, it is enough to express sin C in terms of the sides.

1 1 2 = a2b2 sin2 C = a2b2(1 − cos2 C) 4 4 1 a2 + b2 − c2 2 1 a2 + b2 − c2 a2 + b2 − c2 = a2b2 1 − = a2b2 1 − 1+ 4 2ab 4 2ab 2ab 1 2ab − (a2 + b2 − c2) 2ab +(a2 + b2 − c2) = a2b2 · 4 2ab 2ab 1 1 = [c2 − (a2 − 2ab + b2)][(a2 +2ab + b2) − c2]= [c2 − (a − b)2][(a + b)2 − c2] 16 16 YIU : Trigonometry Workbook 94

1 = [c − (a − b)][c +(a − b)[(a + b) − c][(a + b)+c] 16 1 = (−a + b + c)(a − b + c)(a + b − c)(a + b + c). 16 With a + b + c =2s,wehave

−a + b + c =2(s − a); a − b + c =2(s − b); a + b − c =2(s − c).

From this, 1 2 = · 2(s − a) · 2(s − b) · 2(s − c) · 2s = s(s − a)(s − b)(s − c). 16

6. Three circles of radii r1, r2, r3 are mutually tangent to each other externally. Find the lengths of the sides of the triangle bounded by the external common tangents of the circles.

7. Let OABC be a tetrahedron in which OA = a, OB = b, OC = c,and BOC = α, COA = β, AOB = γ.LetX, Y and P be the projections of C on the lines OA, OB and the plane OAB respectively. (a) Show that the four points O, X, P , Y are concyclic.

OP XY (b) Show that = sin γ . YIU : Trigonometry Workbook 95

8. Suppose three adjacent edges of a tetrahedron have lengths a, b, c, and the angles between them are α (between the edges of lengths b and c), β (between the edges of lengths c and a), and γ (between the edges of lengths a and b). Find the volume of the tetrahedron.

Hints and Comments 1. (i) c =13.53, A =56.33◦, B =63.67◦; (ii) c =31.86, A =30.33◦, B =99.67◦; (iii) c =69.29, A =29.17◦, B =30.83◦;(iv)c =35.67, A =26.77◦, B = 113.23◦;(v)c =48.65, A =21.31◦, B = 113.69◦; 2. ab c A B C (i) 13 14 15 53.13◦ 59.49◦ 67.38◦ (ii) 21 41 50 24.19◦ 53.13◦ 102.7◦ How do you explain the results in (ii) and (iii) (iii) 39 41 50 49.55◦ 53.13◦ 77.32◦ (iv) 25 51 74 9.15◦ 18.92◦ 151.93◦ (v) 25 63 74 18.92◦ 54.82◦ 106.26◦

? 3. Let P and Q be the centers of the circles, and A an intersection. The angle between the tangents at A is the supplement of PAQ. Answer: 81.2◦. t k | k t − t| 4(a). The angle between the two hands at minutes past O’Clock. This is 30( + 60 ) 6 = 11t 11t ◦ |30k − 2 | degrees. The distance between the ends of the two hands is 25 − 24 cos(30k − 2 ) inches. YIU : Trigonometry Workbook 96

Hours 1,7 2,8 3,9 4,10 5,11 6,12 8 6 3 8 5 2 10 7 4 1 Minute 21 11 , 54 11 27 11 0, 32 11 5 11 , 38 11 10 11 , 43 11 16 11 , 49 11 r2r3 5. s = r1+r2+r3; the are of the triangle is = r1r2r3(r1 + r2 + r3). Let α := arctan ; r1(r1+r2+r3 β r3r1 γ r1r2 := arctan r2(r1+r2+r3 ; := arctan r3(r1+r2+r3 . The area bounded between the three circles is 2 2 2 r1r2r3(r1 + r2 + r3) − r1α − r2β − r3γ. √ π 1 For r1 =1,r2 =2andr3 =3,wehaves =6, = 1 · 2 · 3 · 6=6;α =arctan1= 4 ; β =arctan2 ≈; 1 1 1 γ =arctan3 ≈.Area=6− arctan 1 − 4arctan2 − 9arctan3 1 24 1 13 1 13 Remark: 4arctan2 = π − arctan 7 ; 3 arctan 3 =arctan9 ; 9 arctan 3 = 3 arctan 9 = π − 13·37 arctan 27·71 ; 7(a). Since OXP = OY P =90◦, the circle with OP as diameter passes through X and Y . XY 7(b). The circumradius of OXY = 2sinγ . 7(c).

cos2 α +cos2 β − 2cosα cos β cos γ CP2 = c2 − OP 2 = c2(1 − ) sin2 γ c2 = (sin2 γ − cos2 α − cos2 β +2cosα cos β cos γ) sin2 γ c2 = · (1 − (cos2 α +cos2 β +cos2 γ)+2cosα cos β cos γ) sin2 γ 1 ab γ 8. Area of base = 2 sin . The corresponding height is computed in Exercise 7 above. The volume 1 2 2 2 is 6 abc 1 − (cos α +cos β +cos γ)+2cosα cos β cos γ. YIU : Trigonometry Workbook 97

7.5 Stewart Theorem

1. (Stewart Theorem)IfX is a point on the side BC (or its extension) such that BX : XC = λ : µ,then λb2 + µc2 λµa2 AX2 = − . λ + µ (λ + µ)2 (Here, the ratio λ : µ is negative if X lies outside the segment BC).

2(a). If D is the midpoint of BC, then the length of the median AD is given by 1 1 AD2 = (b2 + c2) − a2. 2 4

2(b). Find the lengths of the medians of the triangle whose sides are 136, 170 and 174. (The correct answers are integers). YIU : Trigonometry Workbook 98

3(a). If the bisector of angle A meets BC at X,then a AX2 = bc 1 − ( )2 . b + c

3(b). Find the lengths of the bisectors of the triangle whose sides are 84, 125, 169. (The correct answers are rational numbers).

4. Suppose a = 10, b = 11, and c = 15. Find the height on the side a and the length of the bisector of B. YIU : Trigonometry Workbook 99

5. Let a = 39, b = 57, and c = 21. Find the height on the side a and the median on the side b.

6*. Let a = 1961, b = 1786, and c = 851. Find the lengths of the medians on a and the bisector of B.

Hints and Comments 2(b) ma = 158, mb = 131, mc = 127. t 975 t 26208 t 12600 3(b) a = 7 , b = 253√ , c = 209 . 12 4. Each of these is 5 √21. 15 5. Each of these is 2 √3. 33 6. Each of these is 2 3657. YIU : Trigonometry Workbook 100

7.6 * Area of quadrilateral

Notation Consider a quadrilateral ABCD with lengths of its sides a = AB, b = BC, c = CD, d = DA. Note that these lengths do not determine the shape of the quadrilateral. Suppose the diagonals have lengths p = AC, q = BD, and that the angle between is θ.DenotebyK be area of the quadrilateral.

Exercises 1(a). Construct lines through the vertices of the quadrilateral parallel to its diagonals. Why does the resulting parallelogram have area twice that of the quadrilateral ? What is this in terms of p, q and θ.

1(b). Express p2 in two diﬀerent ways using the cosine formula and show that a2 + b2 − c2 − d2 =2(ab cos B − cd cos D). YIU : Trigonometry Workbook 101

2(a). Show that 2(ab sin B + cd sin D)=4K.

2(b). By computing (ab sin B + cd sin D)2 +(ab cos B − cd cos D)2, show that 16K2 =4(a2b2 + c2d2) − (a2 + b2 − c2 − d2)2 − 8abcd cos(B + D).

Let 2ϕ = B + D. 16K2 =4(a2b2 + c2d2) − (a2 + b2 − c2 − d2)2 − 8abcd(2 cos2 ϕ − 1) =4(a2b2 + c2d2) − (a2 + b2 − c2 − d2)2 +8abcd − 16abcd cos2 ϕ =4(ab + cd)2 − (a2 + b2 − c2 − d2)2 − 16abcd cos2 ϕ =[2(ab + cd) − (a2 + b2 − c2 − d2)][2(ab + cd)+(a2 + b2 − c2 − d2)] − 16abcd cos2 ϕ =[−(a − b)2 +(c + d)2][(a + b)2 − (c − d)2] − 16abcd cos2 ϕ =[(c + d) − (a − b)][(c + d)+(a − b)][(a + b) − (c − d)][(a + b)+(c − d)] − 16abcd cos2 ϕ =(−a + b + c + d)(a − b + c + d)(a + b − c + d)(a + b + c − d) − 16abcd cos2 ϕ. 2(c). Let 2s = a + b + c + d. Show that −a+b+c+d =2(s−a); a−b+c+d =2(s−b); a+b−c+d =2(s−c); a+b+c−d =2(s−d), and that K2 =(s − a)(s − b)(s − c)(s − d) − abcd cos2 ϕ. YIU : Trigonometry Workbook 102

3. (Cyclic quadrilateral) Deduce that if the quadrilateral is cyclic,thenitsareais given by K = (s − a)(s − b)(s − c)(s − d), where s is the semiperimeter of the quadrilateral.

4. What is the condition on four positive numbers a, b, c, d so that there is a quadrilateral whose 4 sides have these lengths in order ? When such a condition is satisﬁed, is italwayspossibletohaveacyclic quadrilateral ?

5. A quadrilateral is said to be circumscribable if its admits an incircle.LetABCD be one such quadrilateral. (a) Show that AB + CD = AD + BC. YIU : Trigonometry Workbook 103

(b) Let s be the semiperimeter. Show that s − a = c, s − b = d, s − c = a,and s − d = a, and deduce that the area of the quadrilateral is √ K = abcd sin ϕ,

ϕ 1 B D where = 2 ( + ).

6. The lengths of the four sides of a cyclic quadrilateral are 4, 3, 5, 6, in order. Show that the quadrilateral is also circumscribable. Find (i) its area, (ii) inradius, (iii) the lengths of the diagonals, and (iv) the circumradius. Chapter 8

Addition formulas and their applications

8.1 Addition formulas

Basic principle The counter – clockwise rotation of the plane about the origin through an angle θ brings the point P (x, y) to the new position Q(x,y)where x cos θ − sin θ x = . y sin θ cos θ y

Since the composite of the rotations through angles θ and ϕ is the rotation through angle θ + ϕ,wehave cos(θ + ϕ) − sin(θ + ϕ) cos θ − sin θ cos ϕ − sin ϕ = sin(θ + ϕ)cos(θ + ϕ) sin θ cos θ sin ϕ cos ϕ

104 YIU : Trigonometry Workbook 105

cos θ cos ϕ − sin θ sin ϕ − sin θ cos ϕ − cos θ sin ϕ = sin θ cos ϕ +cosθ sin ϕ cos θ cos ϕ − sin θ sin ϕ From this, we obtain the addition formulas for the sine and cosine functions:

sin(θ ± ϕ)=sinθ cos ϕ ± cos θ sin ϕ cos(θ ± ϕ)=cosθ cos ϕ ∓ sin θ sin ϕ

Addition formula for the tangent function tan θ ± tan ϕ tan(θ ± ϕ)= . 1 ∓ tan θ tan ϕ Proof.

sin θ cos ϕ±cos θ sin ϕ sin(θ ± ϕ) sin θ cos ϕ ± cos θ sin ϕ θ ϕ tan θ ± tan ϕ tan(θ ± ϕ)= = = cos cos = . cos(θ ± ϕ) cos θ cos ϕ ∓ sin θ sin ϕ cos θ cos ϕ∓sin θ sin ϕ 1 ∓ tan θ tan ϕ cos θ cos ϕ

Example

sin 12◦ =sin(30◦ − 18◦) =sin30◦ cos 18◦ − cos 30◦ sin 18◦ 1 √ √ √ = 10 + 2 5 − 15 + 3) 8 =0.2079116908177593371 ··· cos 12◦ =cos(30◦ − 18◦) = cos 30◦ cos 18◦ +sin30◦ sin 18◦ 1 √ √ = 30 + 6 5+ 5 − 1 8 =0.97814760073380563793 ···

Exercises 1. (Double angle formulas) Show that (a) sin 2θ =2sinθ cos θ; YIU : Trigonometry Workbook 106

(b) cos 2θ =cos2 θ − sin2 θ =2cos2 θ − 1=1− 2sin2 θ; θ 2tanθ (c) tan 2 = 1−tan2 θ .

α α 3 2. Let be an acute angle with tan = 4 . Without using tables, ﬁnd the values of sin 2α, cos 2α, tan 2α.

3. (Triple angle formulas) Show that (a) sin 3θ =3sinθ − 4sin3 θ; (b) cos 3θ =4cos3 θ − 3cosθ; θ 3tanθ−tan3 θ (c) tan 3 = 1−3tan2 θ . YIU : Trigonometry Workbook 107

4(a). Show that

tan θ +tanϕ +tanψ − tan θ tan ϕ tan ψ tan(θ + ϕ + ψ)= . 1 − (tan θ tan ϕ +tanθ tan ψ +tanϕ tan ψ)

4(b). Deduce that in a triangle ABC, (a) tan A +tanB +tanC =tanA tan B tan C; A B B C C A (b) tan 2 tan 2 +tan 2 tan 2 +tan 2 tan 2 =1.

θ ϕ π θ ϕ 5(a). Suppose + = 4 . Show that (1 + tan )(1 + tan )=2. YIU : Trigonometry Workbook 108

π 5(b). Deduce from (a) the value of tan 8 .

1 1 π 6. Show that arctan 1 + arctan 2 + arctan 3 = 2 .

1 1 π 7. Show that 2 arctan 3 + arctan 7 = 4 . YIU : Trigonometry Workbook 109

8.2 Applications

θ θ Trigonometric function of as rational functions of tan 2 t θ Notation := tan 2 .

sin θ cos θ tan θ cot θ sec θ csc θ 2t 1−t2 2t 1−t2 1+t2 1+t2 1+t2 1+t2 1−t2 2t 1−t2 2t

Exercises sin θ t 1. Express 1+cos θ in terms of .

2. Solve the equation 3 sin θ +4cosθ = 2 by writing sin θ and cos θ in terms of t.

Hints and Comments 1. t. 2. More generally, if a sin θ+b cos θ = c,wehave2at+b(1−t2)=c(1+t2), (c+b)t2 −2at+(c−b)=0. 2 2 2 2 There are two realsolutionsfor t if and only if (−2a) − 4(c + b)(c − b) ≥√0, i.e., a + b ≥ c .For 2 2 1 a =3,b =4,c =2,wehave6t − 6t − 2=0,3t +3t − 1=0;6= 6 (−3 ± 21). For another solution, see the next section.

*Solution of a sin θ + b cos θ = c. Given positive numbers a and b, there is a unique YIU : Trigonometry Workbook 110 angle α within 2π such that a b cos α = √ , sin α = √ . a2 + b2 a2 + b2 This helps transform the equation into c sin(θ + α)=√ . a2 + b2

From this, it is clear that the equation is solvable if and only if c2 ≤ a2 + b2,andits solutions are given by c θ = kπ +(−1)k arcsin √ − α, k ∈ Z. a2 + b2

Exercises 1. Solve the following equations

Equation Auxiliary angle α General solution (i) 3 sin θ +4sinθ =2 (ii) 4 sin θ − 3sinθ =1 (iii) 5 sin θ +12sinθ =6 (iv) 12 sin θ − 5sinθ =1

2. What is the largest possible value of 3 sin θ +4cosθ ?

Hints and Comments √ 2. In general, the largest possible value of a sin θ + b cos θ is a2 + b2. YIU : Trigonometry Workbook 111

* An application to the solution of cubic equations Suppose we have a cubic equation in the form x3 + px + q =0. If we write x = r cos θ, this becomes

r3 cos3 θ + pr cos θ = −q.

Since 4 cos3 θ − 3cosθ =cos3θ,ifr is chosen in such a way that

r3 : pr =4:−3, then the equation simpliﬁes into

r3 (4 cos3 θ − 3cosθ)=−q 4 4q cos 3θ = − r3 from which θ, and hence x can be determined. The choice of r being subject to the 3 relation r: pr =4:−3, it is clear that this method requires p to be negative.Inthat r −4p θ 4q 2 case, = 3 . In order to be able to ﬁnd ,(r3 ) should not exceed 1. This means that 4p3 +27q2 must be negative.

3 2 3 Summary If 4p +27q < 0, (which implies p<0), then the cubic equation x +px+q = x r θ r −p 0 has three real roots = cos given by =2 3 ,and q 27 cos 3θ = ···= − · . 2 −p3

Example Solve the equation x3 − 12x +12=0.Here,4p3 +27q2 =122(−48 + 27) = −21 · 122 is negative. We choose r =4andθ according to 3 cos 3θ = ···= − . 4 α 1 − 3 2π α 4π α If := 3 arccos( 4 ), then the other two solutions are 3 + and 3 + .Therootsof the cubic equation are 2kπ x =4cos(α + ),k=0, 1, 2. 3 YIU : Trigonometry Workbook 112

These numbers are 2.76873, −3.88448, and 1.11575.

Exercises. Solve the following cubic equations. x3 + px + q = 0. It is known that each one of them has three real roots.

pqr θ cα 1 cx x x cos 3 = = 3 arccos 1 2 3 (i) −12 8 (ii) −27 2 (iii) −27 40 (iv) −48 12 (v) −48 48

Solution. (i) 3.064, −3.759, 0.6946; (ii) 5.159, −5.233, 0.07409; (iii) 4.173, −5.82, 1.647; (iv) 6.8, −7.05, 0.2503; (v) 6.36, −7.383, 1.022. YIU : Trigonometry Workbook 113

8.3 Sums and products of sines and cosines

1. (Conversion of a product into a sum) Verify the following formulas. 1 sin θ cos ϕ = (sin(θ + ϕ)+sin(θ − ϕ)); 2 1 cos θ cos ϕ = (cos(θ + ϕ)+cos(θ − ϕ)); 2 1 sin θ sin ϕ = − (cos(θ + ϕ) − cos(θ − ϕ)). 2

2. (Conversion of a sum into a product) Verify the following formulas.

θ + ϕ θ − ϕ sin θ +sinϕ =2sin cos ; 2 2 θ + ϕ θ − ϕ sin θ − sin ϕ =2cos sin ; 2 2 θ + ϕ θ − ϕ cos θ +cosϕ =2cos cos ; 2 2 θ + ϕ θ − ϕ cos θ − cos ϕ = −2sin sin . 2 2 YIU : Trigonometry Workbook 114

3(a). Show that sin 20◦ +sin40◦ =sin80◦.

3(b). If sin 10◦ +sin50◦ =sinθ for an acute angle θ,whatisθ ?

◦ ◦ ◦ 1 3(c). Show that cos 20 cos 40 cos 80 = 8 .

θ π θ π 4(a). What is the largest and smallest possible values of cos( + 3 )cos( + 4 )? YIU : Trigonometry Workbook 115

θ θ π 4(b). What is the largest and smallest possible values of cos +cos( + 3 )?

5(a). Show that √ π 3π 5π 7 cos cos cos = . 14 14 14 8

5(b). Show that π 3π 5π 1 π cos +cos +cos = cot . 14 14 14 2 14

6. In ABC, (a) cos2 A +cos2 B +cos2 C =1− 2cosA cos B cos C. YIU : Trigonometry Workbook 116

A B C A B C (b) sin +sin +sin =4cos2 cos 2 cos 2 .

7. Let ABC be an acute–angled triangle and X, Y , Z the midpoints of the sides. A tetrahedron can be formed by folding along XY , YZ and ZX. What is the volume of the tetrahedron ?

8*. Consider the sum

Sn := sin θ +sin2θ + ···+sin(n − 1)θ +sinnθ.

1 Sn θ (a) Multiply each term of by 2 sin 2 and convert the product into a diﬀerence of two cosines.

1 θ·S (b) Combine these terms to write 2 sin 2 n as a diﬀerence of two cosines, and deduce YIU : Trigonometry Workbook 117 that n+1 θ n θ sin 2 sin 2 Sn = . 1 θ sin 2

Cn := cos θ +cos2θ + ···+cos(n − 1)θ +cosnθ.

Hints and Comments 1.Adduptheformulasforsin(θ + ϕ) and sin(θ − ϕ). 2. In the expansion formulas for cos(A + B)andcos(A − B), put A + B = θ and A − B = ϕ. ◦ ◦ ◦ ◦ 1 ◦ ◦ 3(a). sin 20 +sin40 =2sin30 cos 10 =2· 2 · sin 80 =sin80 . 3(b). θ =70◦. 3(c). The easiest way is to multiply the product by 8 sin 20◦ and use the identity 2 sin θ cos θ =sin2θ repeatedly to obtain sin 160◦ =sin20◦. From this the result follows. In the present context, however, you might want to convert the product into a sum. π π 1 7π π 4(a). Since cos(θ + 3 )cos(θ + 4 )= 2 (cos(2θ + 12 )+cos12 , the largest and smallest possible values 1 π ± of the product are 2 (cos 12 1). √ θ θ π θ π π θ π 4(b). cos√ +cos( + 3 )=2cos( + 6 )cos 6 = 3cos( + 6 ). The largest and smallest values of the sum are ± 3. YIU : Trigonometry Workbook 118

√ 1 7. 24 abc cos A cos B cos C. 1 1 1 8(a). 2 sin 2 θ sin kθ =cos(k − 2 )θ − cos(k + 2 )θ. n+1 n 1 1 1 n+1 n sin 2 θ sin 2 θ (b) 2 sin 2 θ · Sn =cos2 θ − cos(n + 2 )θ =2sin 2 θ sin 2 θ.Fromthis,Sn = 1 . sin 2 θ 1 n 1 n 1 1 1 1 (c) 2 sin 2 θ · Cn = j=1 2sin 2 θ cos jθ = j=1 sin(j + 2 )θ − sin(j − 2 )θ =sin(n + 2 )θ − sin 2 θ = n+1 n n+1 n cos 2 θ sin 2 θ 2cos 2 θ sin 2 θ; Cn = 1 . sin 2 θ Chapter 9

The dot and cross products

9.1 The dot product

9.1.1 Vectors in three dimensions We shall denote by 7i the vector joining the origin O to the point (1, 0, 0), and likewise 7j and 7k to (0, 1, 0) and (0, 0, 1) respectively. In this way, the vector joining O to the point P (x, y, z)isx7i + y7j + z7k.

7 7 7 Problem What is the angle between the two vectors v1 := x1i + y1j + z1k and v2 := 7 7 7 x2i + y2j + z2k ?

Solution. Let P =(x1,y1,z1)andQ =(x2,y2,z2) be the endpoints of the two vectors. This is the same as ﬁnding POQ.InPOQ, the lengths of the three sides can be determined easily:

OP 2 v 2 x2 y2 z2 = 1 = 1 + 1 + 1 ; OQ2 v 2 x2 y2 z2 = 2 = 2 + 2 + 2 ; 2 2 2 2 PQ =(x1 − x2) +(y1 − y2) +(z1 − z2) .

By the cosine formula,

2OP · OQ · cos POQ = OP 2 + OP 2 − PQ2 x2 y2 z2 x2 y2 z2 − x − x 2 y − y 2 z − z 2 =(1 + 1 + 1 )+( 2 + 2 + 2 ) [( 1 2) +( 1 2) +( 1 2) ]

119 YIU : Trigonometry Workbook 120

= ···

=2(x1x2 + y1y2 + z1z2).

This suggests deﬁning the dot product of v1 and v2 as

v1 · v2 := x1x2 + y1y2 + z1z2, so that the angle θ between the vectors can be determined easily from the relation

v1 · v2 = v1v2 cos θ.

Basic properties of dot product (1) symmetry: v1 · v2 = v2 · v1. (2) linearity: v · (v1 + v2)=v · v1 + v · v2. (3) v · v = v2. (4) Two vectors are orthogonal (perpendicular) if and only if their dot product is zero. (5) If u is a vector for which u · v =0forall vectors v,thenu =0.

Example. Angles between the diagonals of a parallelepiped A parallelepiped has a vertex at the origin O(0, 0, 0). The three immediate neigbours A, B, C of O are the endpoints of the vectors

OA7 = 7a, OB7 = 7b, OC7 = 7c, the other vertices are YIU : Trigonometry Workbook 121

Vertex opposite to endpoint of vector PO 7a +7b + 7c XA 7b + 7c YB 7a + 7c ZC 7a +7b

The diagonals are the vectors

OP7 = 7a +7b + 7c; AX7 = −7a +7b + 7c; BY7 = 7a −7b + 7c; ZC7 = 7a +7b − 7c.

Exercise 1. Find the angles between each pair of diagonals of the cube with A = (1, 0, 0), B(0, 1, 0), C(0, 0, 1).

OP AX BY ZC OP AX BY ZC

Exercise 2. Find the angles between each pair of diagonals of the parallelepiped with A =(1, 0, 0), B =(1, 1, 0), and C =(1, 1, 1).

OP AX BY ZC OP AX BY ZC YIU : Trigonometry Workbook 122

9.2 The cross product in three dimensions 7 7 7 7 7 7 Let 7x = x1i + x2j + x3k and 7y = y1i + y2j + y3k be two vectors in three dimensions. Their cross product is the vector deﬁned by 7 7 7 7x × 7y := (x2y3 − x3y2)i +(x3y1 − x1y3)j +(x1y2 − x2y1)k.

It is often convenient to express this in the form of a determinant:   7i 7j 7k 7x × 7y   := det  x1 x2 x3  y1 y2 y3

x \ y 7i 7j 7k 7i 0 7k −7j Example 7j −7k 0 7i 7k 7j −7i 0

Basic properties (1) Bilinearity: (i) 7x × (y71 + y72)=7x × y71 + 7x × y72; (ii) (x1 + x2) × 7y = x71 × 7y + x72 × 7y; 7 (iii) (a) × 7y = 7x × (a7y)=a(7x × 7y). (2) 7x × 7x = 70 for every vector x. (3) Skew - symmetry: 7x × 7y = −7y × 7x. (4) If 7x and 7y are linearly dependent, i.e., if one of them is a (scalar) multiple of the other, then 7x × 7y = 70. (5) For arbitrary vectors 7x, 7y,and7z,   x x x  1 2 3  7x · (7y × 7z)=det y1 y2 y3  . z1 z2 z3

(6) The cross product 7x × 7y is perpendicular to both 7x and 7y,(provided7x and 7y are nonzero). Consequently, if 7x and 7y are linearly independent vectors, then 7x × 7y is a normal vector to the plane containing 7x and 7y. YIU : Trigonometry Workbook 123

Exercises 1. Consider a pyramid with a square base on the xy−plane and vertex at (0, 0, 2). Suppose the four corners of the square are (±1, 0, 0) and (0, ±1, 0). (a) Make use of the fact that the angle between a vector 7v and a plane P is the complement of the angle between 7v and the normal of P to ﬁnd the angle between a slant side and the square base.

(b) Make use of the fact that the angle between two planes P is equal to the angle between the normals of the planes to ﬁnd the angle between a slant face and the he square base.

2. The vertices of a tetrahedron are O(0, 0, 0), A(1, 0, 0), B(1, 1, 1), and C(1, 1, 1). Edge OA OB OC AB (a) Complete the table of the vectors representing the edges. Vector 7i 7i + 7j

(b) Find a normal to each face of the tetrahedron.

Opposite face of OABC normal −7k length of normal 1

Opposite face of OABC O A B C

3*. Show that 7x · (7y × 7z)=7y · (7z × 7x)=7z · (7x × 7y).

4(a)*. Verify the identity

x y x y x y 2 x y −x y 2 x y −x y 2 x y −x y 2 x2 x2 x2 y1 y2 y2 . ( 1 1+ 2 2+ 3 3) +( 2 3 3 2) +( 3 1 1 3) +( 1 2 2 1) =( 1+ 2+ 3)( 1+ 2+ 3)

Solution. 2 2 2 2 (x1y1 + x2y2 + x3y3) +(x2y3 − x3y2) +(x3y1 − x1y3) +(x1y2 − x2y1) = = x2 x2 x2 y1 y2 y2 =( 1 + 2 + 3)( 1 + 2 + 3). 4(b)*. Make use of part (a) to show that

(7x · 7y)2 + 7x × 7y2 = x2y2. YIU : Trigonometry Workbook 125

4(c)*. Deduce that 7x × 7y is (numerically) equal to the area of the parallelogram with 7x and 7y as adjacent sides.

4(d)*. If the vertices of a tetrahedron are the endpoints of the vectors 7x, 7y, 7z (including the origin O), then the volume of the tetrahedron is given by   x x x 1  1 2 3  det  y y y  . 6 1 2 3 z1 z2 z3

5. Find the volume of the tetrahedron with vertices (1, 1, 1), (2, 3, 4), (3, 4, 5), and (4, 3, 1).

6. (Setting up an orthonormal frame in a plane) Let 7x and 7y be linearly YIU : Trigonometry Workbook 126 independent vectors. (This means that none of the vectors is a scalar multiple of the other; as such, these vectors are nonzero). 7u 1 x (a) Show that := x is a unit vector. (This is called normalizing a nonzero vector).

(b) Find a scalar a so that 7v := 7y − a7u is orthogonal to 7u. Why must such a vector 7v be nonzero ?

Hints and Comments . 1. The normalto the square base can be taken as k. √ (a) The vector joining the corner (1, 0, 0) to the vertex (0, 0, 2) is −.i +2.k. This has length 5. Since YIU : Trigonometry Workbook 127

−.i .k · .k θ θ √2 θ ( +2 ) = 2, the angle between the slant side and the base is given by cos = 5 , =?. (b) Consider the slant face containing the vertices (0, 0, 2), (1, 0, 0) and (0, 1, 0). The cross product ...... v := (−i +2k) × (−j√+2k)=··· =2i +2j + k (exercise) can be taken as a normalto this face. This normal .v has length 22 +22 +12 =3.Notethat.v · .k = 1. The angle ϕ between this face and the base 1 1 ◦ is given by cos ϕ = 1·3 = 3 ; ϕ =70.53 . 4(c). Let θ be the angle between the vectors .x and .y.Notethat.x · .y = xy cos θ.Itfollowsfrom part (b) that .x × .y2 = x2y2 −x2y2 cos2 θ = x2y2 sin2 θ,and.x × .y = ±xy sinθ. This is the (abolute value of the) parallelogram with .x and .y as adjacent sides. 4(d). Let θ be the angle between .x and .y,andletϕ be the angle between .z and the plane containing 1 .x and .y. If we regard the triangle with edges .x and .y (and area 2 .x × .y) as the base of the tetrahedron, 1 1 1 the corresponding height is z sin ϕ. The volume of the tetrahedron = 3 · 2 .x × .yz sinϕ = 6 .x × .yz cos(90◦ − ϕ). Note that 90◦ − ϕ is the angle between .z and the normal of the base. This latter 1 1 expression is indeed 6 (.x × .y) · .z = ···= 6 .x · (.y × .z). Chapter 10

Spherical Geometry

10.1 Distances on the spherical surface

We study the geometry of the spherical surface.DenotebyS2 the unit sphere in R3, with center at the origin O:

S2 := {(x, y, z) ∈ R3 : x2 + y2 + z2 =1}. A great circle is the intersection of S2 with a plane through O. In spherical geometry, the great circles are analogues of the straight lines in plane geometry.

Distance between two points Let A, B ∈ S2 be two non-antipodal points. There is exactly one great circle passing through them. This is the intersection of S2 with the plane containing the points O, A,andB.Theminor arc AB on this great circle is teh shortest distance between A and B, measured on the sphere. This distance, denoted by d(A, B), is equal to AOB in radian measure. Note that

cos d(A, B)=A · B.

128 YIU : Trigonometry Workbook 129

Exercises 1. Let A be the north pole,andB a point on the equator. What is d(A, B)?

2. What is the distance between the north pole and a point on the latitude 30◦ N?

3. What is the greatest possible distance between two points on S2 ?

Consider the point N(0, 0, 1) ∈ S2 as the north pole, and the great circle through N and E(1, 0, 0) as the standard meridian. Each point P on the surface of the sphere has a colatitude, ϕ := NOP,andifP = ±(0, 0, 1), a longitude θ := EOQ,whereQ is the projection of P on the equatorial plane.

Exercises 4. If P = ±(0, 0, 1) and has longitude θ and colatitude ϕ, then the coordinates of P are (cos θ sin ϕ, sin θ sin ϕ, cos ϕ). YIU : Trigonometry Workbook 130

5. If A and B have spherical coordinates (θ1,ϕ1)and(θ2,ϕ2) respectively, then the distance (measured on the spherical surface) between A and B is given by

cos d(A, B)=cosϕ1 cos ϕ2 +sinϕ1 sin ϕ2 cos(θ1 − θ2).

6. If ABC has B and C on the equator, C the point with longitude 0◦, B with longitude a◦ E. and A on the meridian through C, with latitude b◦ N. What is the distance between A and B ?

A a◦ b◦ B 7. Find the distance between the points with latitude N, longitude 1 W, and b◦ on the equator with longitude 2 W.

8. Assume the earth to be a sphere of radius 4000 miles. Calculate the distance from each of the following places to Singapore, which is on the equator and has longitude 100◦ E. YIU : Trigonometry Workbook 131

Place longitude latitude distance from Singapore (i) Boston 70◦ W40◦ N (ii) San Francisco 125◦ W35◦ N (iii) Hong Kong 110◦ E25◦ N (iv) London 0◦ 50◦ N (v) Johannesburg 28◦ E28◦ S (vi) Rio de Janeiro 42◦ W23◦ S

9. An airplane is ﬂying (at a uniform speed) from above the north pole, maintaining the same height to above a city with latitude a◦ N and longitude b◦ W, along the shortest possible path. What is the latitude of the plane when it is halfway of the ﬂight ?

10. An airplane is ﬂying eastwards (at a uniform speed) from above Singapore (longitude 100 ◦ E and latitude 0 maintaining the same height to above a city with latitude a◦ N and longitude (b + 100)◦ E, along the shortest possible path. What is the latitude of the plane when it is halfway of the ﬂight ?

11. An airplane is ﬂying westwards (at a uniform speed) from above New York (latitude 40◦ N) maintaining the same height to above Beijing with the same latitude, along the shortest possible path. What is the latitude of the plane when it is halfway of the ﬂight ? YIU : Trigonometry Workbook 132

Hints and Comments π π ∗ 1. 2 . More generally, d(A, B)= 2 for every B ∈ lA. π 2. 6 . It is more convenient to talk about the colatitude of a point P as the angle AOP , A being the north pole. For example, points at latitude θ◦ N have colatitude 90 − θ◦, and points at latitude θ◦ S have colatitude 90 + θ◦. 3. π.Notethatifd(A, B)=π, there are inﬁnitely many great circles. Each longitude (meridian) is a great circle passing the north and the south poles. . . . . 5. d(A, B) is the dot product of the vectors cos θ1 sin ϕ1i +sinθ1 sin ϕ1j +cosϕ1k and cos θ2 sin ϕ2i + . . sin θ2 sin ϕ2j +cosϕ2k. π 6. ABC is a right triangle with γ = 2 , BC = a,andAC = b. By the sphericalPythagoras theorem, AB = c is given by cos c =cosa cos b; c = arccos(cos a cos b). 7. c = arccos(cos a cos(b1 − b2)). 8. Distances from Singapore, in miles: Boston 9702.18; San Francisco 8754.31; Hong Kong 1871.41; London 6730.59; Johannesburg 5177.79; Rio de Janeiro 9529.46. 9. 1 (90 + a)◦ since the airplane is ﬂying along the meridian joining the north pole to the destination. 2 ◦ 10. Latitude b Ngivenbysinb =sin45◦ sin b. 11. Latitude b◦ Ngivenbysinb =2sin45◦ sin 40◦. b =65.37◦. YIU : Trigonometry Workbook 133

10.2 Areas of spherical triangles

Consider two lines (minor great circular arcs) intersecting at a point A ∈ S2. The angle α between the two lines is the angle between the planes containing the great circles. To determine this angle, imagine A as the ‘north pole of the globe’. There is a unique ‘equator’ l cutting the given lines at two points B and C. This line l is a common perpendicular to the planes at B and C.Consequently,α = d(A, B). The line l is called ∗ the dual of A, and is denoted by lA. Each pair of lines on S2 bound two antipodal, minor, lunar areas. If α is the angle α · π α between the lines, then each of these lunar regions has area 2π 4 =2 . (Recall that the unit sphere has total area 4π.

Theorem The area of a spherical triangle with angles α, β, γ is equal to the defect from π:

Area of spherical triangles with angles α, β, γ = α + β + γ − π

Exercises 1. Find the area of the spherical triangle bounded by the meridians 0◦ and 30◦ and the great circle cut out by a plane making a 45◦ angle with the xz−plane. Solution. The normals to the planes cutting out the great circles can be chosen as follows. YIU : Trigonometry Workbook 134

meridian 0◦ 30◦ inclined normal 7j

Dual triangles Given a spherical triangle ABC,thereisadual triangle A∗B∗C∗ ∗ ∗ ∗ ∗ ∗ ∗ bounded by the duals lA, lB and lC (so that A is an intersection of the lines lB and lC ). 2. What is the dual of the triangle with vertices the north pole, and the points on the equator with longitudes 0◦ and 45◦ ?

Theorems (1) The dual of A∗B∗C∗ is ABC. (2) If the sides and angles of ABC are a, b, c, α, β, γ respectively, and those of A∗B∗C∗ are a∗,b∗,c∗,α∗,β∗,γ∗ respectively, then

a∗ + α = b∗ + β = c∗ + γ = π; α∗ + a = β∗ + b = γ∗ + c = π.

Exercises 3. Show that similar spherical triangles are congruent. YIU : Trigonometry Workbook 135

Hints and Comments 2 ∗ 1. If we regard a unit normalto a great circle l as a point A ∈ S ,thenl is the dualof A: l = lA. ∗ ∗ Now, the angle between lA and lB is the angle between A and B,namely,d(A, B) = arccos(A · B). It ∗ ∗ ∗ follow that the area bounded by the three lines lA, lB and lC is

arccos(A · B) + arccos(B · C) + arccos(C · A) − π.

For A =.i, B = cos 120◦.i + sin 120◦.j and C = cos 135◦.i + sin 135◦.k,wehave 2 arccos(A · B) = arccos(cos120◦)= π; 3 √ 2 arccos(B · C) = arccos(cos120◦ cos 135◦) = arccos( ); 4 3 arccos(C · A) = arccos(cos135◦)= π. 4 √ √ 2 2 π 3 π − π 2 5 π It follows√ that the area of the spherical triangle is arccos 4 + 3 + 4 = arccos 4 + 12 .(Remark: 2 arccos 4 ≈ 0.385π, so that the area is approximately 0.802π.) .i √1 .i .j Vertex north pole 2 ( + ) 2. Dualequator cut out by yz−plane cut out by vertical plane This has through the bisector of 2nd quadrant. vertices at the north pole, and the points on the equator with longitiudes 90◦ and 135◦.Eachofthese 1 angles has area 4 π. YIU : Trigonometry Workbook 136

10.3 Spherical cosine formula

Let ABC be a spherical triangle with angles α, β, γ and sides a, b, c.

(A) The SAS formulas cos a =cosb cos c +sinb sin c cos α; cos b =cosc cos a +sinc sin a cos α; cos c =cosa cos b +sina sin b cos γ.

(B) The ASA formulas cos α = − cos β cos γ +sinβ sin γ cos a; cos β = − cos γ cos α +sinγ sin α cos b; cos γ = − cos α cos β +sinα sin β cos c.

Exercises γ π 1. (Spherical Pythagoras Theorem) Suppose = 2 .Then cos c =cosa cos b.

2. Find the angles of an equilateral spherical triangle each of whose sides has length π 3 .

Hints and Comments π 2. If a = b = c = 3 ,then cos a − cos b cos c 1 cos α = = ···= . sin b sin c 3 1 ◦ From this, α = arccos 3 =70.53 .