Trigonometry Cram Sheet

August 3, 2016

Contents 6.2 Identities ...... 8

1 Deﬁnition 2 7 Relationships Between Sides and Angles 9 1.1 Extensions to Angles > 90◦ ...... 2 7.1 Law of Sines ...... 9 1.2 The Unit Circle ...... 2 7.2 Law of Cosines ...... 9 1.3 Degrees and Radians ...... 2 7.3 Law of Tangents ...... 9 1.4 Signs and Variations ...... 2 7.4 Law of Cotangents ...... 9 7.5 Mollweide’s Formula ...... 9 2 Properties and General Forms 3 7.6 Stewart’s Theorem ...... 9 2.1 Properties ...... 3 7.7 Angles in Terms of Sides ...... 9 2.1.1 sin x ...... 3 2.1.2 cos x ...... 3 8 Solving Triangles 10 2.1.3 tan x ...... 3 8.1 AAS/ASA Triangle ...... 10 2.1.4 csc x ...... 3 8.2 SAS Triangle ...... 10 2.1.5 sec x ...... 3 8.3 SSS Triangle ...... 10 2.1.6 cot x ...... 3 8.4 SSA Triangle ...... 11 2.2 General Forms of Trigonometric Functions . 3 8.5 Right Triangle ...... 11

3 Identities 4 9 Polar Coordinates 11 3.1 Basic Identities ...... 4 9.1 Properties ...... 11 3.2 Sum and Difference ...... 4 9.2 Coordinate Transformation ...... 11 3.3 Double Angle ...... 4 3.4 Half Angle ...... 4 10 Special Polar Graphs 11 3.5 Multiple Angle ...... 4 10.1 Limaçon of Pascal ...... 12 3.6 Power Reduction ...... 5 10.2 Rose ...... 13 3.7 Product to Sum ...... 5 10.3 Spiral of Archimedes ...... 13 3.8 Sum to Product ...... 5 10.4 Lemniscate of Bernoulli ...... 13 3.9 Linear Combinations ...... 5 10.5 Folium of Descartes ...... 14 3.10 Other Related Identities ...... 5 10.6 Spiral of Fermat ...... 14 3.11 Identities without Variables ...... 6 10.7 Cissoid of Diocles ...... 14 4 Graphs 6 10.8 Epispiral ...... 14 4.1 y = sin x ...... 6 10.9 Lituus ...... 15 4.2 y = cos x ...... 6 10.10 Eight Curve ...... 15 4.3 y = tan x ...... 6 10.11 Butterfly Curve ...... 15 4.4 y = csc x ...... 6 10.12 Strophoid ...... 15 4.5 y = sec x ...... 6 10.13 Cochleoid ...... 16 4.6 y = cot x ...... 6 10.14 Cycloid of Ceva ...... 16 10.15 Freeth’s Nephroid ...... 16 5 Tables 7 5.1 Exact Values of Trigonometric Functions . . 7 11 Miscellaneous Stuff 17 5.2 Relations Between Trig Functions ...... 8 11.1 Pythagorean Triples ...... 17 11.2 Triangle Centers ...... 17 6 Inverse Trigonometric Functions 8 11.3 Area of the Triangle ...... 18 6.1 Principal Values ...... 8 11.4 Other Coordinate Systems ...... 18

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1 Deﬁnition 1.2 The Unit Circle

y Triangle ABC has a right angle at C and sides of length a, b, c. The trigonometric functions of angle A are deﬁned as (0, 1) √ √ follows: 1 3 1 3 − 2 , 2 2 , 2 √ √ √ √ 2 2 2 2 π − 2 , 2 2 , 2 a opposite 2 1. sin A = = 2π π √ 3 3 √ c hypotenuse 3 1 3 1 − , 3π ◦ π , 2 2 4 90 4 2 2 120◦ 60◦ 5π π b adjacent 6 6 2. cos A = = 150◦ 30◦ c hypotenuse (−1, 0) (1, 0) π 180◦ 3600◦◦ 2π x a opposite 3. tan A = = b adjacent 210◦ 330◦ 7π 11π 6 6 240◦ 300◦ √ 5π 7π √ c hypotenuse 3 1 270◦ 3 1 − 2 , − 2 4 4 2 , − 2 4. csc A = = 4π 5π a opposite 3 3 √ √ 3π √ √ 2 2 2 2 2 − 2 , − 2 2 , − 2 √ √ c hypotenuse − 1 , − 3 1 , − 3 5. sec A = = 2 2 2 2 b adjacent (0, −1)

b adjacent 6. cot A = = a opposite

1.3 Degrees and Radians 1.1 Extensions to Angles > 90◦ A radian is that angle θ subtended at center O of a circle by an arc MN equal to the radius r. Since 2π radians = 360◦ A point P in the Cartesian plane has coordinates (x, y), we have: where x is considered as positive along OX and negative 0 0 along OX , while y is considered as positive along OY and 1 radian = 180◦/π = 57.29577951308232 ... ◦ negative along OY . The distance from origin O to point P is positive and denoted by r = px2 + y2. The angle A de- 1◦ = π/180 radians = 0.017453292519943 ... radians scribed counterclockwise from OX is considered positive. If it is described clockwise from OX it is considered negative. 1.4 Signs and Variations For an angle A in any quadrant, the trigonometric functions of A are deﬁned as follows: Quadrant sin A cos A tan A + + + y I 1. sin A = (0, 1) (1, 0) (0, ∞) r + − − II (1, 0) (0, −1) (−∞, 0) x − − + 2. cos A = III r (0, −1) (−1, 0) (0, ∞) − + − y IV 3. tan A = (−1, 0) (0, 1) (−∞, 0) x Quadrant cot A sec A csc A r + + + 4. csc A = I y (∞, 0) (1, ∞) (∞, 1) − − + II r (0, −∞) (∞, −1) (1, ∞) 5. sec A = + − − x III (∞, 0) (−1, ∞) (∞, −1) x − + − 6. cot A = IV y (0, −∞) (∞, 1) (−1, ∞)

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2 Properties and General Forms 2.1.4 csc x

Domain: {x|x 6= kπ, k ∈ }or S (kπ, (k + 1) π) 2.1 Properties Z k∈Z

2.1.1 sin x Range: {y|y ≤ 1 ∪ y ≥ 1} or (−∞, −1] ∪ [1, +∞) Period: 2π Domain: {x|x ∈ R} or (−∞, +∞) Range: {y| − 1 ≤ y ≤ 1} or [−1, 1] VA: x = kπ where k ∈ Z Period: 2π x-intercepts: none

VA: none Parity: odd x-intercepts: kπ where k ∈ Z Parity: odd 2.1.5 sec x

π S (k−1)π (k+1)π Domain: x|x 6= + kπ, k ∈ Z or , 2.1.2 cos x 2 k∈Z 2 2 Range: {y|y ≤ 1 ∪ y ≥ 1} or (−∞, −1] ∪ [1, +∞) Domain: {x|x ∈ R} or (−∞, +∞) Range: {y| − 1 ≤ y ≤ 1} or [−1, 1] Period: 2π

π Period: 2π VA: x = 2 + kπ where k ∈ Z VA: none x-intercepts: none x-intercepts: π + kπ where k ∈ 2 Z Parity: even Parity: even 2.1.6 cot x 2.1.3 tan x S Domain: {x|x 6= kπ, k ∈ Z} or k∈ (kπ, (k + 1) π) Domain: x|x 6= π + kπ, k ∈ or S (k−1)π , (k+1)π Z 2 Z k∈Z 2 2 Range: {y|y ∈ R} or (−∞, +∞) Range: {y|y ∈ R} or (−∞, +∞) Period: π Period: π π VA: x = kπ where k ∈ Z VA: x = 2 + kπ where k ∈ Z x-intercepts: π + kπ where k ∈ x-intercepts: kπ where k ∈ Z 2 Z Parity: odd Parity: odd

2.2 General Forms of Trigonometric Functions

Given some trigonometric function f (x), its general form is represented as y = Af (B (x − C)) + D, where its amplitude 2π π is |A|, its period is |B| or |B| (for tangent and cotangent), its phase shift is C, and its vertical translation is D units upward (if D > 0) or D units downward (if D < 0). The maximum and minimum value for sin x and cos x is A + D and −A + D respectively.

amplitude period phase shift vertical translation

1 1 1 1 f f f f

−1 0 1 2 3 −1 0 1 2 3 −1 0 1 2 3 −1 0 1 2 3 a −1 a −1 a −1 a −1

−2 −2 −2 −2

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3 Identities 3.2 Sum and Diﬀerence

3.1 Basic Identities sin (α ± β) = sin α cos β ± cos α sin β

Reciprocal Identities cos (α ± β) = cos α cos β ∓ sin α sin β 1 1 tan α ± tan β csc θ = ; sin θ = tan (α ± β) = sin θ csc θ 1 ∓ tan α tan β 1 1 cot α cot β ∓ 1 sec θ = ; cos θ = cot (α ± β) = cos θ sec θ cot β ± cot α 1 1 cot θ = ; tan θ = tan θ cot θ 3.3 Double Angle sin θ csc θ = cos θ sec θ = tan θ cot θ = 1 sin 2α = 2 sin α cos α

Ratio Identities cos 2α = cos2 α − sin2 α = 1 − 2 sin2 α = 2 cos2 α − 1 sin θ sin θ tan θ = ; cos θ = ; sin θ = cos θ tan θ 2 tan α cos θ tan θ tan 2α = 1 − tan2 α cos θ cos θ cot θ = ; sin θ = ; cos θ = sin θ cot θ sin θ cot θ 3.4 Half Angle

Pythagorean Identities Let Qn, where n ∈ {1, 2, 3, 4}, denote the set of all angles within the nth quadrant of the Cartesian plane. sin2 θ + cos2 θ = 1; sin2 θ = 1 − cos2 θ; cos2 θ = 1 − sin2 θ

2 2 2 2 2 2 q 1−cos α α tan θ + 1 = sec θ; tan θ = sec θ − 1; sec θ − tan θ = 1 α  2 if 2 ∈ (Q1 ∪ Q2) sin = q 2 2 2 2 2 2 2 1−cos α α cot θ + 1 = csc θ; cot θ = csc θ − 1; csc θ − cot θ = 1 − 2 if 2 ∈ (Q3 ∪ Q4)

q 1+cos α α Co-function Identities α  2 if 2 ∈ (Q1 ∪ Q4) cos = q π 2 − 1+cos α if α ∈ (Q ∪ Q ) sin − θ = cos θ 2 2 2 3 2 π α sin α 1 − cos α cos − θ = sin θ tan = = = csc α − cot α 2 2 1 + cos α sin α π tan − θ = cot θ 2 3.5 Multiple Angle π csc − θ = sec θ sin 3α = 3 sin α − 4 sin3 α 2 π 3 sec − θ = csc θ cos 3α = 4 cos α − 3 cos α 2 3 tan α − tan3 α π tan 3α = cot − θ = tan θ 2 2 1 − 3 tan α sin 4α = 4 sin α cos α − 8 sin3 α cos α Parity Identities cos 4α = 8 cos4 α − 8 cos2 α + 1 sin (−A) = − sin A 4 tan α − 4 tan3 α cos (−A) = cos A tan 4α = 1 − 6 tan2 α + tan4 α tan (−A) = − tan A n X n (n − i) π sin (nα) = cosi α sinn−i α sin csc (−A) = − csc A i 2 i=0 sec (−A) = sec A n X n (n − i) π cos (nα) = cosi α sinn−i α cos i 2 cot (−A) = − cot A i=0

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3.6 Power Reduction Definition 1 − cos 2θ sin2 θ = The two-argument form of the arctangent function, denoted 2 by tan−1 (y, x) gathers information on the signs of the in- puts in order to return the appropriate quadrant of the com- 1 + cos 2θ cos2 θ = puted angle. Thus, it is defined as: 2 tan−1 y if x > 0,  x  −1 y 2 1 − cos 2θ tan + π if x < 0 and y ≥ 0, tan θ =  x 1 + cos 2θ  −1 y −1 tan x − π if x < 0 and y < 0, tan (y, x) = π + 2 if x = 0 and y > 0, 3 3 sin θ − sin 3θ  sin θ = − π if x = 0 and y < 0, 4  2 undefined if x = 0 and y = 0. 3 cos θ + cos 3θ cos3 θ = 4 Sine and Cosine 3 − 4 cos 2θ + cos 4θ sin4 θ = In the case of a non-zero linear combination of a sine and 8 cosine wave (which is just a sine wave with a phase shift of π 3 + 4 cos 2θ + cos 4θ 2 ), we have: cos4 θ = 8 a sin x + b cos x = c sin (x + θ) 10 sin θ − 5 sin 3θ + sin 5θ √ sin5 θ = where c = ± a2 + b2 and θ satisfies the equations c cos θ = 16 a and c sin θ = b, or θ = tan−1 (b, a). 10 cos θ + 5 cos 3θ + cos 5θ cos5 θ = 16 Arbitrary Phase Shift

More generally, for an arbitrary phase shift, we have: 3.7 Product to Sum a sin x + b sin (x + θ) = c sin (x + ϕ) 1 sin α cos β = [sin (α + β) + sin (α − β)] √ 2 where c = ± a2 + b2 + 2ab cos θ, and ϕ satisﬁes the equations c cos ϕ = a + b cos θ and c sin ϕ = b sin θ or ϕ = 1 cos α sin β = [sin (α + β) − sin (α − β)] tan−1 (b sin θ, a + b cos θ). 2 1 cos α cos β = [cos (α + β) + cos (α − β)] 3.10 Other Related Identities 2 • If x + y + z = π, then sin 2x + sin 2y + sin 2z = 1 sin α sin β = [cos (α + β) − cos (α − β)] 4 sin x sin y sin z. 2 • Triple Tangent Identity. If x + y + z = π, then tan x + tan y + tan z = tan x tan y tan z. 3.8 Sum to Product π • Triple Cotangent Identity. If x + y + z = 2 , then θ ± ϕ θ ∓ ϕ sin θ ± sin ϕ = 2 sin cos cot x + cot y + cot z = cot x cot y cot z. 2 2 • Ptolemy’s Theorem. If w + x + y + z = π, then θ + ϕ θ − ϕ sin (w + x) sin (x + y) = sin w sin y + sin x sin z. cos θ + cos ϕ = 2 cos cos 2 2 • cot x cot y + cot y cot z + cot z cot x = 1 √ θ + ϕ θ − ϕ −1 cos θ − cos ϕ = −2 sin sin • a cos x + b sin x = a2 + b2 cos x − tan (b, a) 2 2 α+β sin α+sin β • Tangent of an Average. tan 2 = cos α+cos β = cos α−cos β 3.9 Linear Combinations − sin α−sin β

x π For some purposes it is important to know that any linear • tan x + sec x = tan 2 + 4 combination of sine waves of the same period or frequency n (n+1)α nα X sin sin ϕ + but diﬀerent phase shifts is also a sine wave with the same • sin (ϕ + iα) = 2 2 sin α period or frequency, but a diﬀerent phase shift. i=0 2

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n (n+1)α nα 1 X sin cos ϕ + ◦ ◦ ◦ ◦ • cos (ϕ + iα) = 2 2 • cos 24 + cos 48 + cos 96 + cos 168 = sin α 2 i=0 2 ∞ n • cos 2π + cos 2 · 2π + cos 4 · 2π + cos 5 · 2π + X Y mπ 21 21 21 21 • cos = 1 cos 8 · 2π + cos 10 · 2π = 1 2n + 1 21 21 2 n=1 m=1 π √ • cos = cos 36◦ = 1 ( 5 + 1) = 1 ϕ 3.11 Identities without Variables 5 4 2

◦ ◦ ◦ 1 π √ • Morrie’s Law. cos 20 · cos 40 · cos 80 = • sin = sin 18◦ = 1 ( 5 − 1) = 1 ϕ−1 8 10 4 2 √ 3 • sin 20◦ · sin 40◦ · sin 80◦ = 2 ◦ 2 ◦ 2 ◦ 8 • sin 18 + sin 30 = sin 36

4 Graphs 4.1 y = sin x 4.4 y = csc x

2 2

1 1

−2π 3π −π − π π π 3π 2π −2π 3π −π − π π π 3π 2π − 2 2 2 2 − 2 2 2 2 −1 −1

−2 −2

4.2 y = cos x 4.5 y = sec x

2 2

1 1

−2π 3π −π − π π π 3π 2π −2π 3π −π − π π π 3π 2π − 2 2 2 2 − 2 2 2 2 −1 −1

−2 −2

4.3 y = tan x 4.6 y = cot x

2 2

1 1

−2π 3π −π − π π π 3π 2π −2π 3π −π − π π π 3π 2π − 2 2 2 2 − 2 2 2 2 −1 −1

−2 −2

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5 Tables

5.1 Exact Values of Trigonometric Functions

A◦ A rad sin A cos A tan A cot A sec A csc A

0◦ 0 0 1 0 ∞ 1 ∞ √ √ √ √ √ √ √ √ √ √ ◦ 1 1 15 π/12 4 6 − 2 4 6 + 2 2 − 3 2 + 3 6 − 2 6 + 2 √ √ √ √ ◦ 1 1 1 2 30 π/6 2 2 3 3 3 3 3 3 2 √ √ √ √ ◦ 1 1 45 π/4 2 2 2 2 1 1 2 2 √ √ √ √ ◦ 1 1 1 2 60 π/3 2 3 2 3 3 3 2 3 3 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 75 5π/12 4 6 + 2 4 6 − 2 2 + 3 2 − 3 6 + 2 6 − 2

90◦ π/2 1 0 ±∞ 0 ±∞ 1 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 105 7π/12 4 6 + 2 − 4 6 − 2 − 2 + 3 − 2 − 3 − 6 + 2 6 − 2 √ √ √ √ ◦ 1 1 1 2 120 2π/3 2 3 2 − 3 − 3 3 −2 3 3 √ √ √ √ ◦ 1 1 135 3π/4 2 2 − 2 2 −1 −1 − 2 2 √ √ √ √ ◦ 1 1 1 2 150 5π/6 2 − 2 3 − 3 3 − 3 − 3 3 2 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 165 11π/12 4 6 − 2 − 4 6 + 2 − 2 − 3 − 2 + 3 − 6 − 2 6 + 2

180◦ π 0 −1 0 ∓∞ −1 ±∞ √ √ √ √ √ √ √ √ √ √ ◦ 1 1 195 13π/12 − 4 6 − 2 − 4 6 + 2 2 − 3 2 + 3 − 6 − 2 − 6 + 2 √ √ √ √ ◦ 1 1 1 2 210 7π/6 2 − 2 3 3 3 3 − 3 3 −2 √ √ √ √ ◦ 1 1 225 5π/4 − 2 2 − 2 2 1 1 − 2 − 2 √ √ √ √ ◦ 1 1 1 2 240 4π/3 − 2 3 − 2 3 3 3 −2 − 3 3 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 255 17π/12 − 4 6 + 2 − 4 6 − 2 2 + 3 2 − 3 − 6 + 2 − 6 − 2

270◦ 3π/2 −1 0 ±∞ 0 ∓∞ −1 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 285 19π/12 − 4 6 + 2 4 6 − 2 − 2 + 3 − 2 − 3 6 + 2 − 6 − 2 √ √ √ √ ◦ 1 1 1 2 300 5π/3 − 2 3 2 − 3 − 3 3 2 − 3 3 √ √ √ √ ◦ 1 1 315 7π/4 − 2 2 2 2 −1 −1 2 − 2 √ √ √ √ ◦ 1 1 1 2 330 11π/6 − 2 2 3 − 3 3 − 3 3 3 −2 √ √ √ √ √ √ √ √ √ √ ◦ 1 1 345 23π/12 − 4 6 − 2 4 6 + 2 − 2 + 3 − 2 + 3 6 − 2 − 6 + 2

360◦ 2π 0 1 0 ∓∞ 1 ∓∞

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5.2 Relations Between Trig Functions

sin θ = u cos θ = u tan θ = u csc θ = u sec θ = u cot θ = u √ √ u 1 u2 − 1 1 sin θ u 1 − u2 √ √ 1 + u2 u u 1 + u2 √ √ 1 u2 − 1 1 u cos θ 1 − u2 u √ √ 1 + u2 u u 1 + u2 √ u 1 − u2 1 √ 1 tan θ √ u √ u2 − 1 1 − u2 u u2 − 1 u √ 1 1 1 + u2 u √ csc θ √ u √ 1 + u2 u 1 − u2 u u2 − 1 √ 1 1 √ u 1 + u2 sec θ √ 1 + u2 √ u 1 − u2 u u2 − 1 u √ 1 − u2 u 1 √ 1 cot θ √ u2 − 1 √ u u 1 − u2 u u2 − 1

6 Inverse Trigonometric Functions Reciprocal Identities 1 −1 sin−1 = csc−1 x If x = sin y, then y = sin x, i.e. the angle whose sine is x x or arcsine of x, is a multiple-valued function of x which is a 1 collection of single-valued functions called branches. Simi- cos−1 = sec−1 x larly, the other inverse trigonometric functions are multiple- x valued. 1 csc−1 = sin−1 x For many purposes, a particular branch is required. This is x called the principal branch and the values for this branch 1 sec−1 = cos−1 x are called principal values. x ( 1 π − tan−1 x = cot−1 x if x > 0 tan−1 = 2 x π −1 −1 6.1 Principal Values − 2 − tan x = cot x − π if x < 0 ( Since none of the six trigonometric functions are one-to-one, 1 π − cot−1 x = tan−1 x if x > 0 cot−1 = 2 they are restricted in order to have inverse functions. There- x 3π −1 −1 2 − cot x = π + tan x if x < 0 fore the ranges of the inverse functions are proper subsets of the domains of the original functions. Negative Identities

−1 −1 Principal values for x ≥ 0 Principal values for x < 0 sin (−x) = − sin x −1 −1 −1 π π −1 cos (−x) = π − cos x 0 ≤ sin x ≤ 2 − 2 ≤ sin x < 0 −1 π π −1 tan−1 (−x) = − tan−1 x 0 ≤ cos x ≤ 2 2 < cos x ≤ π −1 π π −1 csc−1 (−x) = − csc−1 x 0 ≤ tan x < 2 − 2 < tan x < 0 −1 π π −1 −1 −1 0 < cot x ≤ 2 2 < cot x < π sec (−x) = π − sec x −1 π π −1 −1 −1 0 ≤ sec x < 2 2 < sec x ≤ π cot (−x) = π − cot x −1 π π −1 0 < csc x ≤ 2 − 2 ≤ csc x < 0 Complementary Identities π sin−1 x + cos−1 x = 6.2 Identities 2 π tan−1 x + cot−1 x = In all cases it is assumed that principal values are used. 2

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π sec−1 x + csc−1 x = 7.4 Law of Cotangents 2 a+b+c Let s be the semiperimeter, that is, s = 2 , and r be Sum and Diﬀerence Identities the radius of the inscribed circle, then:

p p α β γ sin−1 α ± sin−1 β = sin−1 α 1 − β2 ± β 1 − α2 cot cot cot 1 2 = 2 = 2 = s − a s − b s − c r cos−1 α ± cos−1 β = cos−1 αβ ∓ p(1 − α2) (1 − β2) and furthermore that the inradius is given by: −1 −1 −1 α ± β tan α ± tan β = tan r 1 ∓ αβ (s − a)(s − b)(s − c) r= s 7 Relationships Between Sides and 7.5 Mollweide’s Formula Angles Each of these identities uses all six parts of the triangle—the 7.1 Law of Sines three angles and the lengths of the three sides. α−β α−β a + b cos 2 a − b sin 2 a b c = γ ; = γ = = c sin c cos sin α sin β sin γ 2 2 7.6 Stewart’s Theorem Extended Law of Sines

A γ

b c d b a a R C n m B α D c β

Let D be a point in BC of 4ABC. If |BD| = m, |CD| = n, and |AD| = d, then b2m + c2n = a d2 + mn. a b c = = = 2R, sin α sin β sin γ 7.7 Angles in Terms of Sides where R is the circumradius of the triangle. a+b+c Let s = 2 be the semiperimeter of the triangle, then:

2 7.2 Law of Cosines α = sin−1 ps (s − a)(s − b)(s − c) bc b2 + c2 − a2 p cos α = ; a = b2 + c2 − 2bc cos α 2 2bc β = sin−1 ps (s − a)(s − b)(s − c) ac a2 + c2 − b2 p cos β = ; b = a2 + c2 − 2ac cos β 2 2ac γ = sin−1 ps (s − a)(s − b)(s − c) ab a2 + b2 − c2 p cos γ = ; c = a2 + b2 − 2ab cos γ 2ab

7.3 Law of Tangents

1 a − b tan 2 (α − β) = 1 a + b tan 2 (α + β)

1 b − c tan 2 (β − γ) = 1 b + c tan 2 (β + γ)

1 c − a tan 2 (γ − α) = 1 c + a tan 2 (γ + α)

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8 Solving Triangles The known characteristics are the side c and the angles α, β. The third angle γ = 180◦ − α − β. A general form triangle has six main characteristics: three Two unknown side can be calculated from the law of sines: linear (side lengths a, b, c) and three angular (α, β, γ). The classical plane trigonometry problem is to specify three of c sin α c sin β a = ; b = . the six characteristics and determine the other three. A tri- sin γ sin γ angle can be uniquely determined in this sense when given any of the following: The procedure for solving an AAS triangle is same as that for an ASA triangle: First, ﬁnd the third angle by using the • Three sides (SSS) angle sum property of a triangle, then ﬁnd the other two sides using the law of sines. • Two sides and the included angle (SAS)

• Two sides and an angle not included between them (SSA), if the side length adjacent to the angle is 8.2 SAS Triangle shorter than the other side length.

• A side and the two angles adjacent to it (ASA)

• A side, the angle opposite to it and an angle adjacent to it (AAS).

For all cases in the plane, at least one of the side lengths must be specified. If only the angles are given, the side lengths cannot be determined, because any similar triangle is a solution. Here the lengths of sides a, b and the angle γ between these sides are known. The third side can be determined from the law of cosines: Notes p • To find an unknown angle, the law of cosines is safer c = a2 + b2 − 2ab cos γ than the law of sines. The reason is that the value of sine for the angle of the triangle does not uniquely Now we use law of cosines to find the second angle: determine this angle. For example, if sin β = 0.5, the angle β can be equal either 30◦ or 150◦. Using the b2 + c2 − a2 law of cosines avoids this problem: within the inter- α = cos−1 val from 0◦ to 180◦ the cosine value unambiguously 2bc determines its angle. On the other hand, if the an- Finally, β = 180◦ − α − γ. gle is small (or close to 180◦), then it is more robust numerically to determine it from its sine than its cosine because the arc-cosine function has a divergent derivative at 1 (or −1). 8.3 SSS Triangle

• We assume that the relative position of specified characteristics is known. If not, the mirror reflection of the triangle will also be a solution. For example, three side lengths uniquely define either a triangle or its reflection.

8.1 AAS/ASA Triangle

Let three side lengths a, b, c be speciﬁed. To ﬁnd the angles α, β, the law of cosines can be used:

b2 + c2 − a2 a2 + c2 − b2 α = cos−1 ; β = cos−1 . 2bc 2ac

Then angle γ = 180◦ − α − β.

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8.4 SSA Triangle 9 Polar Coordinates

A point P can be located by rectangular coordinates (x, y) or polar coordinates (r, θ). The angle θ is a directed angle, that is, it is positive if it is measured counterclockwise from the initial side to the terminal side, and negative if it is measured clockwise. The value r is a directed distance, it is positive if the point P lies on the terminal side of θ and negative if P is on the extension of the terminal side. This case is not solvable in all cases; a solution is guaranteed to be unique only if the side length adjacent to the angle is shorter than the other side length. Assume that two sides 9.1 Properties b, c and the angle β are known. The equation for the angle γ can be implied from the law of sines: • Every ordered pair of polar coordinates (r, θ) locates a unique point in the plane. c sin β sin γ = b • However, a point P on the plane may be specified by an infinite number of ordered pairs (r, θ). We denote further D = c sin β (equation’s right side). There b • The pole O may be specified by the ordered pair (0, θ) are four possible cases: where θ ∈ R. 1. If D > 1, no such triangle exists because the side b • Let P (r, θ) be a point in the polar plane. Then does not reach line BC. For the same reason a solu- (r, θ + 2kπ) are also coordinates of the point P for tion does not exist if the angle β ≥ 90◦ and b ≤ c. any k ∈ Z. ◦ 2. If D = 1, a unique solution exists: γ = 90 , i.e., the • It can also be shown that ((−1)n r, θ + nπ) are also triangle is right-angled. coordinates of P , where n ∈ Z. 3. If D < 1, two alternatives are possible. 9.2 Coordinate Transformation

Polar to Rectangular ( x = r cos θ y = r sin θ

(a) If b < c, the angle γ may be acute: γ = sin−1 D or obtuse: γ0 = 180◦ − γ. The picture above Rectangular to Polar shows the point C, the side b and the angle γ as ( p 2 2 the ﬁrst solution, and the point C0, side b0 and r = x + y 0 −1 −1 y the angle γ as the second solution. θ = tan (y, x) ≈ tan x (b) If b ≥ c then β ≥ γ (the larger side corresponds where tan−1 (y, x) is the two-argument form of the arctan- to a larger angle). Since no triangle can have two gent function (see section 3.9). obtuse angles, γ is acute angle and the solution γ = sin−1 D is unique.

Once γ is obtained, the third angle α = 180◦ − β − γ . 10 Special Polar Graphs The third side can then be found from the law of sines: Theorem b sin α a = A polar graph is: sin β 1. symmetric with respect to the polar axis if an 8.5 Right Triangle equivalent equation is obtained when (r, θ) is replaced by either (r, −θ) or (−r, π − θ). Solving right triangles is simply using the deﬁnitions of the π trigonometric functions and the Pythagorean theorem to de- 2. symmetric with respect to the 2 -axis if an equiv- termine the other parts. The right angle γ = 90◦ is always alent equation is obtained when (r, θ) is replaced by assumed to be given. either (r, π − θ) or (−r, −θ).

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a 3. symmetric with respect to the pole if an equiv- • Dimpled limaçon if 1 < b < 2 alent equation is obtained when (r, θ) is replaced by π 2 either (−r, θ) or (r, π + θ). 2π π 3 3

10.1 Limaçon of Pascal 5π π 6 6 A polar equation of the form r = a+b cos θ or r = a+b sin θ has a polar graph which is called a limaçon. Let OQ be a line joining origin O to any point in Q on a 0 2 4 circle of diameter b passing through O. Then the curve is π the locus of all points in P such that |PQ| = a

Types of Limaçons 7π 11π 6 6 a • Looped limaçon if 0 < b < 1 π 2 4π 5π 2π π 3 3 3 3 3π 2

a 5π π • Convex limaçon if b ≥ 2 6 6 π 2 2π π 3 3

0 2 4 6 π 5π π 6 6

7π 11π 0 2 4 6 6 6 π

4π 5π 3 3 3π 2 7π 11π 6 6 a • Cardioid if b = 1 π 2 4π 5π 2π π 3 3 3π 3 3 2

5π π Symmetry and Direction 6 6 For a > 0 and b > 0:

0 1 2 3 4 π symmetry points to the r = a + b cos θ polar axis right r = a − b cos θ polar axis left 7π 11π r = a + b sin θ π -axis top 6 6 2 π r = a − b sin θ 2 -axis bottom

4π 5π 3 3 3π 2

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10.2 Rose • A rose with an even number of leaves is symmetric π with respect to the polar axis, the 2 -axis, and the A rose with n leaves has a polar equation r = a cos (nθ) or pole. r = a sin (nθ) where a is a constant and n is an odd integer.

π 10.3 Spiral of Archimedes 2 2π π 3 3 The polar graph of a polar equation r = aθ where θ > 0 and a ∈ R is called a spiral. 5π π 6 6 π 2 2π π 3 3

0 2 4 π 5π π 6 6

7π 11π 0 2 4 6 6 6 π

4π 5π 3 3 3π 2 7π 11π 6 6 For an even integer n, the polar graph of an equation r = a cos (nθ) or r = a sin (nθ) is a rose with 2n leaves. 4π 5π 3 3 3π π 2 2 2π π 3 3 10.4 Lemniscate of Bernoulli

5π π 2 2 6 6 A polar equation r = a cos 2θ or r = a sin 2θ has a polar graph that is called a lemniscate.

π 2 0 0.5 1 1.5 2 2π π π 3 3

5π π 6 6

7π 11π 6 6

0 1 2 3 4 4π 5π π 3 3 3π 2

Properties 7π 11π 6 6 • The length of one leaf in the polar graph of a rose is |a|. 4π 5π 3 3 • If n is odd, then the graph of the polar equation 3π 2 r = a cos (nθ) is symmetric with respect to the polar axis. • If n is odd, then the graph of the polar equation π r = a sin (nθ) is symmetric with respect to the 2 -axis.

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10.5 Folium of Descartes and center (a, 0) with the extension of OP . Then the cissoid of Diocles is the curve which satisﬁes OP = RS. It has A folium is a plane curve proposed by Descartes to chal- a polar equation r = 2a sin θ tan θ. lenge Fermat’s extremum-ﬁnding techniques. It has a polar 3a sec θ tan θ equation r = 3 . π 1+tan θ 2 2π π 3 3 π 2 2π π 3 3 5π π 6 6

5π π 6 6

0 2 4 π 0 1 2 3 4 π

7π 11π 6 6

7π 11π 6 6 4π 5π 3 3 3π 2 4π 5π 3 3 3π 2 10.8 Epispiral 10.6 Spiral of Fermat The epispiral is a plane curve with a polar equation r = a sec (nθ). Then there are n sections if n is odd (in blue), or The Fermat’s spiral, also known as the parabolic spiral, has 2n sections if n is even (in red). A slightly more symmetric a polar equation r2 = a2θ. The resulting spiral is symmetric version considers instead r = a |sec (nθ)|. with respect to the origin. π 2 π 2π π 2 3 2π π 3 3 3

5π π 5π π 6 6 6 6

0 2 4 6 0 1 2 π π

7π 11π 6 7π 11π 6 6 6

4π 5π 3 3 4π 5π 3π 3 3 3π 2 2

10.7 Cissoid of Diocles

Given an origin O and a point P on the curve, let S be the point where the extension of the line OP intersects the line x = 2a and R be the intersection of the circle of radius a

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10.9 Lituus 10.11 Butterﬂy Curve

The lituus is the locus of the point P moving such that The butterﬂy curve is a transcendental plane curve discov- the area of a circular sector remains constant. It means a ered by Temple H. Fay. The curve is given by the polar sin θ 5 1 “crook,” in the sense of a bishop’s crosier. It has a polar equation r = e − 2 cos 4θ + sin 24 (2θ − π) . equation r2θ = a2. π 2 2π π π 3 3 2 2π π 3 3

5π π 6 6 5π π 6 6

0 1 2 3 4 π 0 0.5 1 1.5 2 π

7π 11π 6 6 7π 11π 6 6

4π 5π 3 3 3π 4π 5π 2 3 3 3π 2 10.12 Strophoid 10.10 Eight Curve Let C be a curve, let O be a fixed point (the pole), and let O0 be a second fixed point. Let P and P 0 be points on a line The eight curve, also known as the lemniscate of Gerono, through O meeting C at Q such that P 0Q = QP = QO0. is given by the polar equation r2 = a2 sec4 θ cos (2θ). It The locus of P and P 0 is called the strophoid of C with has vertical tangents (±a, 0) and horizontal tangents at respect to the pole O and fixed point O0. Its polar equation √ b sin(a−2θ) 2 1 is r = . ± 2 a, ± 2 a . sin(a−θ)

π π 2 2π π 2 2π π 3 3 3 3

5π π 5π π 6 6 6 6

0 5 10 0 1 2 3 4 π π

7π 11π 7π 11π 6 6 6 6

4π 5π 3 3 4π 5π 3π 3 3 3π 2 2

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10.13 Cochleoid 10.15 Freeth’s Nephroid

A cochleoid is a snail-shaped curve similar to a strophoid It is a strophoid of a circle with the pole O at the center of a sin θ which can be represented by the polar equation r = θ . the circle and the ﬁxed point P on the circumference of the θ circle. It has a polar equation r = a 1 + 2 sin 2 . π 2 2π π π 3 3 2 2π π 3 3

5π π 6 6 5π π 6 6

0 1 2 3 π 0 1 2 3 π

7π 11π 6 6 7π 11π 6 6

4π 5π 3 3 3π 4π 5π 2 3 3 3π 2 10.14 Cycloid of Ceva

The cycloid of Ceva is a polar curve that can be used for angle trisection. It has the polar equation r = 1 + 2 cos 2θ.

π 2 2π π 3 3

5π π 6 6

0 1 2 3 π

7π 11π 6 6

4π 5π 3 3 3π 2

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11 Miscellaneous Stuﬀ Incenter

11.1 Pythagorean Triples The incenter I is the center of the incircle for a triangle. The corresponding radius of the incircle is known as the inradius. A Pythagorean triple (a, b, c) is a triple of positive inte- gers such that a2 + b2 = c2 . It is primitive if the great- The incenter can be constructed as the intersection of angle est common divisor of a, b, and c is 1. For any primi- bisectors. It is also the interior point for which distances to tive Pythagorean triple, either m or n is even, but not the sides of the triangle are equal. both (i.e. m 6≡ n mod 2). When that is done, then ev- For a triangle with Cartesian vertices (x1, y1), (x2, y2), ery primitive Pythagorean triple (a, b, c) is of the form (x3, y3), the Cartesian coordinates of the incenter are given (a, b, c) = m2 − n2, 2mn, m2 + n2 where m and n are rel- by: atively prime (i.e. gcd (m, n) = 1) and 1 ≤ n < m. ax1 + bx2 + cx3 ay1 + by2 + cy3 m n a b c m n a b c I (x0, y0) = , 2 1 3 4 5 11 8 57 176 185 a + b + c a + b + c 3 2 5 12 13 11 10 21 220 221 4 1 15 8 17 12 1 143 24 145 Orthocenter 4 3 7 24 25 12 3 135 72 153 5 2 21 20 29 12 5 119 120 169 The intersection H of the three altitudes AHA, BHB, and 5 4 9 40 41 12 7 95 168 193 CHC of a triangle is called the orthocenter. 6 1 35 12 37 12 9 63 216 225 For a triangle with Cartesian vertices (x1, y1), (x2, y2), 6 3 27 36 45 12 11 23 264 265 (x3, y3), the Cartesian coordinates of the circumcenter are 6 5 11 60 61 13 2 165 52 173 given by: 7 2 45 28 53 13 4 153 104 185 7 4 33 56 65 13 6 133 156 205 x1tA + x2tB + x3tC y1tA + y2tB + y3tC H (x0, y0) = , 7 6 13 84 85 13 8 105 208 233 tA + tB + tC tA + tB + tC 8 1 63 16 65 13 10 69 260 269 where t , t , t is equal to tan A, tan B, tan C respectively. 8 3 55 48 73 13 12 25 312 313 A B C 8 5 39 80 89 14 1 195 28 197 8 7 15 112 113 14 3 187 84 205 Circumcenter 9 2 77 36 85 14 5 171 140 221 9 4 65 72 97 14 7 147 196 245 The circumcenter is the center O of a triangle’s circumcir- 9 6 45 108 117 14 9 115 252 277 cle. It can be found as the intersection of the perpendicular 9 8 17 144 145 14 11 75 308 317 bisectors. 10 1 99 20 101 14 13 27 364 365 For a triangle with Cartesian vertices (x1, y1), (x2, y2), 10 3 91 60 109 15 2 221 60 229 (x3, y3), the Cartesian coordinates of the circumcenter are 10 5 75 100 125 15 4 209 120 241 given by: 10 7 51 140 149 15 6 189 180 261 10 9 19 180 181 15 8 161 240 289 O (x0, y0) = (Ox,Oy) 11 2 117 44 125 15 10 125 300 325 (y −y ) x2+(y −y )(y −y ) +x2(y −y )+x2(y −y ) 15 12 81 360 369 2 3 ( 1 1 2 1 3 ) 2 3 1 3 1 2 11 4 105 88 137 where Ox = 2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2)) 11 6 85 132 157 15 14 29 420 421 2 2 2 2 2 2 2 2 2(x2−x3)(−x1+x3−y1 +y3 )+2(x1−x3)(x2−x3+y2 −y3 ) and Oy = . 4(x1(y2−y3)+x2(y3−y1)+x3(y1−y2)) 11.2 Triangle Centers Excenter Centroid An excenter, denoted Ji, is the center of an excircle of a triangle. An excircle is a circle tangent to the extensions of The geometric centroid (center of mass) of the polygon ver- two sides and the third side. It is also known as an escribed tices of a triangle is the point G which is also the intersec- circle. tion of the triangle’s three triangle medians. The point is therefore sometimes called the median point. The centroid is always in the interior of the triangle.

For a triangle with Cartesian vertices (x1, y1), (x2, y2), (x3, y3), the Cartesian coordinates of the centroid are given by:

x + x + x y + y + y G (x , y ) = 1 2 3 , 1 2 3 0 0 3 3

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11.3 Area of the Triangle Equilateral Triangle √ Base and Altitude 3 2 A4 = a 1 4 A = bh 4 2 where b is the base and h is the altitude. Isosceles Triangle These are special cases of the formulas given above. SAS Triangle b p 1 1 1 1 A = 4a2 − b2 = a2 sin θ A = ab sin γ = bc sin α = ac sin β 4 4 2 4 2 2 2

SSS Triangle Pick’s Theorem 1 A = I + B − 1 Given three sides a, b, and c, the area of the triangle can be 4 2 determined by Heron’s formula: where I is the number of internal lattice points and B is the number of lattice points lying on the border of the triangle. p A4 = s (s − a)(s − b)(s − c) 1 = p4a2b2 − (a2 + b2 − c2)2 11.4 Other Coordinate Systems 4

a+b+c Let a, b, c denote the side length and A, B, C denote the where s = 2 is the semiperimeter. The area can also be determined by: angles at the respective vertices.

abc A = Barycentric Coordinates 4 4R where R is the circumradius. centroid G 1 : 1 : 1 orthocenter H tan A : tan B : tan C incenter I sin A : sin B : sin C AAS Triangle circumcenter O sin 2A : sin 2B : sin 2C a2 sin β sin γ b2 sin α sin γ c2 sin α sin β excenters Ja,Jb,Jc {−a : b : c, a : −b : c, a : b : −c} A = = = 4 2 sin α 2 sin β 2 sin γ where the missing angle can be easily determined through Trilinear Coordinates ◦ α + β + γ = 180 . centroid G csc A : csc B : csc C orthocenter H sec A : sec B : sec C ASA Triangle incenter I 1 : 1 : 1 circumcenter O cos A : cos B : cos C a2 b2 c2 A = = = excenters Ja,Jb,Jc {−1 : 1 : 1, 1 : −1 : 1, 1 : 1 : −1} 4 2 (cot β + cot γ) 2 (cot α + cot γ) 2 (cot α + cot β)

Three Vertices

Given the vertices of the triangle (xi, yi) ∀i ∈ {1, 2, 3}, the shoelace formula can be applied for n = 3 as follows:

2 2 1 X X A = x y + x y − x y − x y 4 2 i i+1 3 1 i+1 i 1 3 i=1 i=1 1 = |x y + x y + x y − x y − x y − x y | 2 1 2 2 3 3 1 2 1 3 2 1 3 However, if one of the vertices is at the origin, then the area of the triangle can be simpliﬁed:

1 x2 x3 1 A4 = = |x2y3 − x3y2| 2 y2 y3 2 where k·k denotes the absolute value of the determinant.