# Thread: Simple expression for trig matrix cubed

1. ## Simple expression for trig matrix cubed

Does anyone know how to determine a simple expression for $\displaystyle A^3$, where A is the following matrix?

cos@ sin@
-sin@ cos@

*I used @ as theta

I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!

2. Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $\displaystyle A^3$, where A is the following matrix?

cos@ sin@
-sin@ cos@

*I used @ as theta

I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
A is the standard rotation matrix by angle @. A^3 = A.A.A => rotate by 3@. Therefore A^3 = .....

3. ## Reply

Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $\displaystyle A^3$, where A is the following matrix?

cos@ sin@
-sin@ cos@

*I used @ as theta

I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
Please see attached file

4. ## Thank you!

You have helped me so much. There is nothing in my text for linear algebra that looks anything like this matrix. I wasn't sure how to simplify the sines and cosines. Thanks again. Your work is very clear!

5. Hello, jennifer1004!

We are expected to know some basic identities:

. . $\displaystyle \begin{array}{cccccc}\cos^2\!x-\sin^2\!x \:=\:\cos2x & & \cos x\cos y - \sin x\sin y \:=\: \cos(x+y) \\ 2\sin x\cos x \:=\: \sin2x & & \sin x\cos y + \cos x\sin y \:=\:\sin(x+y) \end{array}$

Given: .$\displaystyle A \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$

Find $\displaystyle A^3$

$\displaystyle A^2 \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$ .$\displaystyle = \;\begin{bmatrix}\cos^2\!\theta-\sin^2\!\theta & 2\sin\theta\cos\theta \\ \text{-}2\sin\theta\cos\theta & \cos^2\!\theta -\sin^2\!\theta \end{bmatrix}$ .$\displaystyle = \;\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta\end{bmatrix}$

$\displaystyle A^3 \;=\;\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta \end{bmatrix}$

. . $\displaystyle = \;\begin{bmatrix}\cos\theta\cos2\theta - \sin\theta\sin2\theta & \sin2\theta\cos\theta + \sin\theta\cos2\theta \\ \text{-}\sin\theta\cos2\theta - \sin2\theta\cos\theta & \cos\theta\cos2\theta - \sin\theta\sin2\theta \end{bmatrix}$

. . $\displaystyle = \;\begin{bmatrix}\cos3\theta & \sin3\theta \\ \text{-}\sin3\theta & \cos3\theta \end{bmatrix}$

6. ## Thanks Soroban!

Those identities are what I was looking for. I wasn't sure how to simplify. Thanks for sharing!

### cube of a matrix in cos sin - sin cos

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