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?

use the $a\sin(2x)+b\cos(2x)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin(2x)+\frac{b}{\sqrt{a^2+b^2}}\cos(2x)\right)$
and set $\cos(\phi)=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$
