I am asked as a part of a question to express $\sin(2x)-8\cos(2x)$ as a single sine function.
I know it has something to do with the trigonometric identity $\sin(a-b)=\sin(a) \cos(b)-\cos(a)\sin(b)$ but I can't get my head around it because of that $8$ in front of $\cos2x$.
Any tips on how I can move on?