Calculate the matrices $[{A{π \over 4}]}^v$ for all possible values of $v$, when $A(\varphi)=\left(\begin{matrix} \cos\varphi &\sin\varphi\\ -\sin\varphi & \cos\varphi\end{matrix}\right)$ .
So $A\left(\frac{\pi}{4}\right)=\left(\begin{matrix} \frac{\sqrt{2}}{2} &\frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{matrix}\right)$. I tried to calculate $ A^v$ for $v=[0,4]$, $v\in Z$ with the hope that it would be a periodical function but this method wasn't fruitful.
*Note:*The question doesn't state that $ v\in Z$ but I guess it should.
I suspect there was a typo in the OP (couldn't tell until the formatting got fixed). Surely it should be $A(\phi)=\binom{\ \cos\phi\,\, -\sin\phi}{\sin \phi\,\,\cos\phi}$. Assuming so:
Hint: This is a funny case where calculating $A(\phi)^\nu$ is easier than $A(\pi/4)^\nu$, because you can see the pattern easier.
In fact the thing to do is this: Calculate the product $A(\phi)A(\theta)$. Now review the sum formulas for sine and cosine, and use them to rewrite your answer for $A(\phi)A(\theta)$. The answer to the original question should now be clear...