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if $M(x)=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}$ I have started solving it but I don't know if I should go the long way by multiplying each element of $M(x)$, $M(y)$ respectively with each other.

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    The assignment definitely means the standard matrix product, not the component-wise one. While there is a very precise geometric idea behind that identity, the proof is just a direct calculation (of course you need to know the elementary trigonometry to write $\cos(x+y)$ and $\sin(x+y)$ in terms of $\sin x,\ \cos x,\ \sin y,\ \cos y$). – 2017-01-30
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    This is matrix of rotation in the plane... see https://en.wikipedia.org/wiki/Rotation_matrix#In_two_dimensions.. – 2017-01-30

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Hint :$$M(x)=\begin{bmatrix}cosx&-sinx\\sinx&cosx\end{bmatrix}\\ M(y)=\begin{bmatrix}cosy&-siny\\siny&cosy\end{bmatrix}\\ \begin{bmatrix}cosx&-sinx\\sinx&cosx\end{bmatrix}\begin{bmatrix}cosy&-siny\\siny&cosy\end{bmatrix}=\\ \begin{bmatrix}cosx cosy-sinxsiny &-sinxcosy-cosxsiny\\sinxcosy+cosxsiny&cosxcosy-sinxsiny\end{bmatrix}=\\\begin{bmatrix}cos(x+y)&-sin(x+y)\\sin(x+y)&cos(x+y)\end{bmatrix} $$ You need to use $$\color{red} {sin(x\pm y)=sinx cosy \pm cos x siny\\cos(x+y)=cosxcosy -sinx siny \\cos(x-y)=cos x cos y +sinx sin y }$$