Download Rotation Matrix ($)

Rotation Matrix
Suppose that a
.
We let
—R
2
R: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
a
clockwise
rotates the
plane counterclockwise. If a < 0. then R is
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
(1,0)
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(oct)
( $)
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).
0 )
g
(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)
Using some of tile trigonometric
know that
Cosine.
cos (a
idleiltities
+
we
sin (a
learned in
+
the chapter Sine
wTe
sin (a +
[Leinnia (8)]
= cos(a)
and
cos a
sin(a +
)
[Lemma (9)]
[Lemma (10)]
= —sin(a)
Therefore,
R(0, 1)
(
cos (a +
sin (a +
236
))
=
(— sin(a), cos(a))
and
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
That’s what tins theorem says.
an angle of a +
.
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2
We know what the rotation function R :
in polar coordinates. The formula is
does to vectors written
r(cos(O + ),sin(O +
R(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)
R(c(a,b))
=
is a vector,
then
cR(a,b)
(c())
I
j)
b)
1
c
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 +
cR(a, b)
), sin(O + ))
Lemma (15). If (a, b) and (c, d) are two vectors, then
R((a,b) + (c,d))
=
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Ra(a,b
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)
sm()
R(),
sin(c)
cos()
—
, amid the matrix above are the same function,
0
Proof: To show that R
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)
sin() cos(a)
(‘
—
1
(cos(c) —sin(c) (a
b
sin() cos()
)
—
—
=
(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.
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(a,b)
1
R
R((a,O) + (O,b))
= R(a. 0) + R(0, b)
= R(a(1O)) +R(b(O,1))
= aR(1, 0) + bR(0, 1)
= a( cos(a), sin(a)) + b(
Lemrria (15)
Lemma (14)
—
Examples
sin(), cos(c))
= (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.
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Angle sum formulas
Theorenm(16 allows us to write
Theorem (13) tells us that Ra+
this equation with matrices
—sin(a+/3)
sin(c + /3) cos( + /3)
(cos(+/3)
(cos() cos(/3)
—
—
cos(c) —sin(c)
sin(c) cos()
sin(cr) sin(/3)
—
sin() cos(/3) + cos() sin(/3)
—
cos(ci) sin($)
cos(/3) —sin(/3)
sin(/3) cos(/3)
—
sin() cos(/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:
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cos(a + B)
=
cos(a) cos(/3)
sin(a + 3)
=
sin(a) cos(8) + cos(a) sin(/3)
—
siu(a) sin(B)
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)
sin(2a)
2
cos(a)
—
2
sin(a)
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
= RR
; Lemma (9) in the matrix equation
1
_ = RR; Lemma (10) in the matrix equation R
6
R
RR; Lemma
+
8
(11) and (12) in the equation R
9
Ri’; and the period identities for sine
and cosine in the equation R
?-.
2
9
2 =R
+
9
o. conic.
Let’s rotate the hyperbola xy = clockwise by an angle of
that we will apply the rotation matrix R_. to the hyperbola.
.
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211
That means,
To find the new equation for our rotated hyperbola, we’ll precompose the
-with the the in
equation of the original hyperbolatIie equation xy
verse of R. Notice that
/cos() -sin(
1=I
4
sin()
cos())
and replaces y with
is the function that replaces x with
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.
-x
2
2
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+ y.
Exercises
For most of these exercises you’ll need to use the rotation matrix
R
—
—
(cos(c) —sin(c)
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
For #5-8, use the matrices
in Cartesian coordinates,
5.) R(3, 5)
4.) R
2
3.) R
2.) R
and R from above to write the given vectors
8.) R(5, —9)
7.) R(—2, —7)
6.) R(4, —8)
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).
11.) Use an angle sum formula to find sin
(i).
12.) Suppose that (cos(O), sin(O))
formulas to find (cos(20),sin(29)).
=
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(Hint:
(, ).
f
+
(—).)
Use the double angle
13.) Find an equation for the hyperbola xy
angle of
=
—
rotated clockwise by an
4
2 = 0 is a parabola. Find an equation for the
14.) The solution of y x
Write your answer in
parabola rotated counterclockwise by an angle of
2 + Bxy + Cy
the form x
2 + Dx + Ey = 0.
—
.
13
For #15-17, find the solutions of the given equations.
+x1E1
15.) 1
16.) ln(x) + 2 ln(x)
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=
7
17.) (x + 1)2
16