Suppose that a . We let R:R2—R be the function defiled as follows: Aiiy vector in the plane can be written in polar coordmates as r(cos(9), sin(O)) where r > 0 and 0 R. For any such vector, we define ( r( cos(0), sin(0))) = r( cos(0 ± a), sin(0 + a)) Notice that the function Ra doesn’t change the norms of vectors (the num ber r), it just affects their direction, which is measured by the unit circle coordinate. We call the function Rc, rotation of the plane by angle a. 5 ‘ (cs),s .
%
c co ($)
If a > 0. then R rotates the plane counterclockwise. If a < 0. then R is a clockwise rotation.
Examples.
. R(1, 0) is the point (1, 0) rotated by an angle of a, which is the point (cos(a), sin(a)) (cos(Sn (oct)
(1,0)
235 R(0, 1) is the point (0, 1) rotated by au angle of a. Or alternatively, we could rotate the point (1,0) by an angle of to get to the point (0, 1), and then we could rotate another a to get to the point R (0, 1).
g0 (o, ) (i,o)
Written as an equation, the previous sentence is
R(O, 1) — R(H (1, 0) Using the previous example, we have that
R(0. 1) R+(1. 0) cos (a + sin (a +
Using some of tile trigonometric idleiltities we learned in the chapter Sine and
Cosine. wTe know that
sin (a + = cos(a) [Leinnia (8)] and cos a sin(a + ) [Lemma (9)] = —sin(a) [Lemma (10)]
Therefore, R(0, 1) ( cos (a + sin (a + ) ) = (— sin(a), cos(a))
236 . If a > 0, then Ra is a counterclockwise rotation of angle a. The function R is a clockwise rotation by a, If first we rotate the plane coun terclockwise by angle a, and then we rotate the plane clockwise by angle a it will be as if we had done nothing at all to the plane. In other words, Ra and R are inverse functions. That is
= R
0
C’
Theorem (13). R o R Proof: If first we rotate the plane by an angle of B, and then we rotate the plane by an angle of a, what we would have done is rotated the plane by an angle of a + . That’s what tins theorem says.
* * * * * * * * * * * * *
237 We know what the rotation function R : 2 does to vectors written in polar coordinates. The formula is R(r(cos(O),sin(O))) r(cos(O + ),sin(O + )) What’s less clear is what the formula for Ra should be for vectors written in Cartesian coordinates. For example, what’s R(3, 7)? We’ll answer this question below, in Theorem 16. Before that though, we need a couple of more lemmas.
Lemma (14). If c is a scalar and (a, b) is a vector, then
R(c(a,b)) = cR(a,b) (c())
Ic1b) j)
Proof: We can write (a, b) as vector in polar coordinates. Let’s say that (a,b) = r(cos(O),sin(O)). Then
R(c(a,b)) = R(cr(cos(0),sin(O))) = cr( cos(O + ), sin(O + )) cR(a, b)
Lemma (15). If (a, b) and (c, d) are two vectors, then
R((a,b) + (c,d)) = Ra(a,b
238 Proof: In the first picture are the vectors (a, b), (c, d), and their sum (a, b) + (c, d). The second picture shows these three vectors rotated by an angle of : the vectors Ra(a, b), R(c, d), and R((a, b) + (c, d)). Notice in tins second picture that R((a, b) + (c, d)) would be the result of adding the vectors Ra(a, b) and R(c, d), which is what the lemma claims to be true.
Now we are ready to describe the rotation function R using Cartesian coordmates. The functioll can be written as a matrix function, and we know how matrix functions affect vectors written in Cartesian coordinates.
Theorem (16). Ra 2 2 is the same function as the matrix function
(cos() —sin(c) siii() cos()
For short,
(cos(c) — sin(c) R(), sm() cos()
Proof: To show that R0, amid the matrix above are the same function, we’ll input the vector (a, b) into each function and check that we get the same output. First let’s check the matrix function. Writing out vectors interchangeably as row vectors and column vectors we have that
(cos() —sin(c) ( — (cos(c) —sin(c) (a sin() cos(a) ‘ 1 sin() cos() ) b
— (acos() — bsin()
— asin() +bcos(a)
= (acos() — bsin(), asin() + bcos())
Now let’s check that we’d get the same output if we input the vector (a, b) into the function Ra. 239 R1(a,b) R((a,O) + (O,b)) = R(a. 0) + R(0, b) = R(a(1O)) +R(b(O,1)) Lemrria (15) = aR(1, 0) + bR(0, 1) Lemma (14)
= a( cos(a), sin(a)) + b( — sin(), cos(c)) Examples = (acos(),asin()) + (—bsin(),bcos(a))
= (a cos() — b sin(), a sin() + b cos())
Notice that the matrix arid the function R(- gave us the same output, so they are the same function. N ( / Z * * * * * * * *
Angle sum formulas Theorem (13) tells us that Ra+ Theorenm(16 allows us to write this equation with matrices
—sin(a+/3) — cos(c) —sin(c) cos(/3) —sin(/3) (cos(+/3)sin(c + /3) cos( + /3) sin(c) cos() sin(/3) cos(/3)
(cos() cos(/3) — sin(cr) sin(/3) — cos(ci) sin($) — sin() cos(/3)
sin() cos(/3) + cos() sin(/3) — sin() sin(s) + cos(a) cos(fl)
The top left entry of the matrix we started with has to equal the top left entry of the matrix that we ended with, since tire two matrices are equal. Similarly, the bottom left entries are equal. That gives us two equations two trigonometric identities that are known as the angle sum formulas:
240 cos(a + B) = cos(a) cos(/3) — siu(a) sin(B)
sin(a + 3) = sin(a) cos(8) + cos(a) sin(/3)
Double angle formulas
If we write the angle sum formulas with a = 8 then w&d have two more trigonometric identities, called the double angle formulas:
cos(2a) cos(a)2 — sin(a)2
sin(2a) 2 sin(a) cos(a)
More identities encoded in matrix multiplication The angle sum and double angle formulas are encoded in matrix multipli cation. as we saw above. In fact all but one of the trigonometric identities that weve see so far are encoded iii matrix multiplication. Lemma (8) can be seen in the matrix equation = RR1; Lemma (9) in the matrix equation R6_ = RR; Lemma (10) in the matrix equation R8+ RR; Lemma (11) and (12) in the equation R9 Ri’; and the period identities for sine and cosine in the equation R9+2 = R92?-. o. conic.
Let’s rotate the hyperbola xy = clockwise by an angle of . That means, that we will apply the rotation matrix R_. to the hyperbola. ;9
211 To find the new equation for our rotated hyperbola, we’ll precompose the equation of the original hyperbolatIie equation xy -with the the in verse of R. Notice that /cos() -sin(
4 1=I sin() cos())
is the function that replaces x with — and replaces y with + y. Therefore, the equation for the rotated hyperbola is
/1 1 ‘/1 1 ‘, 1
Using the distributive law, the above equation is the same as 19 121 2’ Multiplying both sides by 2 leaves us with the equivalent equation 9 9 x-—y =1 Tins is the equation for the rotated hyperbola.
2 2 -x /
7
/ N / /
242 Exercises
For most of these exercises you’ll need to use the rotation matrix
— (cos(c) —sin(c) R — sin() cos(a)
For #1-4 use the values of sin() and cos() that you know to write the given matrices using exact numbers that do not involve trigonometric func tions such as sin or cos. For example, as we saw when rotating the hyperbola at the end of the chapter,
R=I i \‘
1.) R 2.) R 3.) R 4.) R2
For #5-8, use the matrices and R from above to write the given vectors in Cartesian coordinates,
5.) R(3, 5) 6.) R(4, —8) 7.) R(—2, —7) 8.) R(5, —9)
9.) Use a trigonometric identity from a previous chapter the find the de terminailt of the matrix R.
10.) Use an angle suni formula to find cos (i). (Hint: f + (—).) 11.) Use an angle sum formula to find sin (i).
12.) Suppose that (cos(O), sin(O)) Use the double angle = (,). formulas to find (cos(20),sin(29)). 243 13.) Find an equation for the hyperbola xy = — rotated clockwise by an angle of
4
14.) The solution of y — x2 = 0 is a parabola. Find an equation for the parabola rotated counterclockwise by an angle of . Write your answer in the form x2 + Bxy + Cy2 + Dx + Ey = 0.
13
For #15-17, find the solutions of the given equations.
15.) 1+x1E1 16.) ln(x) + 2 ln(x) = 7 17.) (x + 1)2 16
244