If $y=-e^x\cos2x$, show that $\frac{d^2y}{dx^2}=5e^x\sin(2x+\tan^{-1}(\frac{3}{4}))$ 
If $y=-e^x\cos2x$, show that $\frac{d^2y}{dx^2}=5e^x\sin(2x+\alpha)$ where $\alpha=\tan^{-1}(\frac{3}{4})$.

I've managed to figure out that
$$
\frac{d^2y}{dx^2}=e^x(4\sin2x+3\cos2x)
$$
But I'm not sure how I can massage it into the form above. Wolfram Alpha lists it as an alternative form but doesn't show how you get there - what do I need to do and which trig identities are relevant here?

$4\sin2x+3\cos2x=5(\frac{4}{5}\sin2x+\frac{3}{5}\cos2x)$
Note we want $\frac{4}{5}\sin2x+\frac{3}{5}\cos2x=\sin(2x+\alpha)$ for some $\alpha$.
By compound angle formula, we know $$\sin(2x+\alpha)=\sin2x\cos(\alpha)+\cos2x\sin(\alpha)$$
So in order to fulfill the requirement, we only need to set $$\cos(\alpha)=\frac{4}{5},\sin(\alpha)=\frac{3}{5}$$
Such $\alpha$ exists.
Hence $5(\frac{4}{5}\sin2x+\frac{3}{5}\cos2x)=5\sin(2x+\alpha)$ with $\cos(\alpha)=\frac{4}{5},\sin(\alpha)=\frac{3}{5}$, that is $\tan(\alpha)=\frac{3}{4}$.
