Show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 50. Exercise 3: Suppose that $\alpha$ is some angle. If $a=4\cos^3\alpha-3\cos\alpha$ and $b=3\sin\alpha-4\sin^3\alpha$, show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$
The attempt at a solution:
In order to show that, I understand that I have to show that $a^2+b^2=1$, now i have expanded $(4\cos^3\alpha-3\cos\alpha)^2+(3\sin\alpha-4\sin^3\alpha)^2$, but all I get is messy expression. I would appreciate some hints, thank you in advance.

By De Moivre's formula, we have that 
$$\begin{align}
\cos3x+i\sin3x&=(\cos x+i\sin x)^3
\\&=\cos^3x+3i\cos^2x\sin x-3\sin^2x\cos x-i\sin^3 x
\\&=\cos^3x-3(1-\cos^2x)\cos x+i[-\sin^3x+3(1-\sin^2x)\sin x]
\\&=4\cos^3x-3\cos x+i[3\sin x-4\sin^3 x]
\end{align}$$
which implies
$$\cos3x=4\cos^3x-3\cos x\,\,\,\,\,\,\,\,\,\,\,\,\text{ and }\,\,\,\,\,\,\,\,\,\,\,\,\sin 3x=3\sin x-4\sin^3x$$
This just about solves it (let $\theta=3\alpha$).
