# Linear Algebra and Trig Identity Proof

I am working on the following question. It involves finding a proof for a trig identity using linear algebra. The proof is one involving $sin(\alpha +\theta)$ and $cos(\alpha +\theta)$, as you will see. I will go through where I am up to, progressing through each part of the question.

Let $T_{\alpha}$ be the linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ which is the rotation counterclockwiseby $\alpha$, and $T_{\theta}$ the counterclockwise rotation by θ.

(A) Write down the standard matrices for $T_{\alpha}$ and $T_{\theta}$, explain your reasoning.

Let the standard matrix for $T_{\alpha}$ be $A$ and let the standard matrix for $T_{\theta}$ be $B$, then

$$A=\begin{bmatrix} \cos (\alpha) & -\sin (\alpha) \\ \sin (\alpha) & \cos (\alpha) \\ \end{bmatrix}$$

and

$$B=\begin{bmatrix} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \\ \end{bmatrix}$$

On to the next part of the queston.

(B) Explain what the linear transformation $T_{\alpha} \circ T_{\theta}$ does to $\mathbb{R}^2$.

It first rotates a given point by $\alpha$ degrees and then rotates the given point by $\theta$ degrees.

(C) Compute the matrix for $T_{\alpha} \circ T_{\theta}$ by multiplying the matrices for $T_{\alpha}$ and $T_{\theta}$

So,

$$AB = \begin{bmatrix} \cos (\alpha)\cos (\theta) - \sin (\alpha) \sin (\theta) & -\cos (\alpha)\sin (\theta) - \sin (\alpha) \cos (\theta) \\ \sin (\alpha)\cos (\theta) + \cos (\alpha) \cos (\theta) & -\sin (\alpha)\sin (\theta) + \cos (\alpha) \cos (\theta) \\ \end{bmatrix}$$

(D) On the other hand, from the description in part (b), you can directly write down the matrix for $T_{\alpha} \circ T_{\theta}$. What is that matrix?

If that matrix is $C$, then

$$C=\begin{bmatrix} \cos (\alpha + \theta) & -\sin (\alpha + \theta) \\ \sin (\alpha + \theta) & \cos (\alpha + \theta) \\ \end{bmatrix}$$

(E) Since the matrices from parts (c) and (d) are describe the same linear transformation, they must be equal. What identities among sin and cos must therefore be true?

So I must set $AB=C$. Then

$$\begin{bmatrix} \cos (\alpha)\cos (\theta) - \sin (\alpha) \sin (\theta) & -\cos (\alpha)\sin (\theta) - \sin (\alpha) \cos (\theta) \\ \sin (\alpha)\cos (\theta) + \cos (\alpha) \cos (\theta) & -\sin (\alpha)\sin (\theta) + \cos (\alpha) \cos (\theta) \\ \end{bmatrix}=\begin{bmatrix} \cos (\alpha + \theta) & -\sin (\alpha + \theta) \\ \sin (\alpha + \theta) & \cos (\alpha + \theta) \\ \end{bmatrix}$$

These are the identities I was looking for! Now the next part has me worried.

Using a similar idea, ﬁnd formulas for $\sin(3\theta)$ and $\cos(3\theta)$ in terms of $\sin(\theta)$ and $\cos(\theta)$.

Now I am not completely hopeless - I was able to come up with this next bit. But I am not sure if I have done things correctly.

I'll use the transformation $T_{\theta}$ from before. The standard matrix is

$$A=\begin{bmatrix} \cos (\alpha) & -\sin (\alpha) \\ \sin (\alpha) & \cos (\alpha) \\ \end{bmatrix}$$

I thought that maybe if I transformed a point three times it would rotate it $3\theta$ degrees, i.e with a standard matrix $AAA$. The result was $$AAA=B$$ $$=$$$$\begin{bmatrix} \cos^{3}(\theta)-\cos(\theta)\sin^2(\theta)-2\cos(\theta)\sin^2(\theta) & -\sin(\theta)\cos^2(\theta)+\sin^3(\theta)-2\cos^2(\theta)\sin(\theta) \\ 2\sin(\theta)\cos^2(\theta)-\sin^3(\theta)+\cos^2(\theta)\sin(\theta) & -2\sin^3(\theta)\cos(\theta)-\sin^2(\theta)cos(\theta)+\cos^3(\theta)\ \\ \end{bmatrix}$$

As before I would say that $AAA$ is equal to a transformation matrix

$$A'=\begin{bmatrix} \cos (3\alpha) & -\sin (3\alpha) \\ \sin (3\alpha) & \cos (3\alpha) \\ \end{bmatrix}$$

Then setting $A'=B$ would result in some identities. They just seem quite long and I am not sure if what I have done is correct or checks out. Any help would be appreciated.

-
Yes, this is correct. – wj32 Oct 29 '12 at 5:40
Looks good. Thank you for showing all your work, it is nice to get a question which has some real effort put into it. A very minor nitpick. $T_\alpha \circ T_\theta$ is given by $T_\alpha(T_\theta(x))$ so it rotates by $\theta$ and then $\alpha$. The order is irrelevant in this case, but it is important for other transformations. – EuYu Oct 29 '12 at 5:48

• For (B) EuYu already noted that $T_\alpha\circ T_\theta$ is first rotation by $\theta$ then by $\alpha$. For rotations it doesn't matter since the composition in each direction is the rotation by $\alpha+\theta$. In case of other linear maps that will matter.
• For (F) this seems to be what is asked for. Note that you could simplify these expressions a bit using other trigonometric identities, e.g. $\sin^2\alpha+\cos^2\alpha=1$.