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.
 A: First of all:
$$a = 4\cos^3 \alpha - 3\cos \alpha = 4\cos\alpha (1-\sin^2\alpha) - 3\cos\alpha = \cos \alpha -4\cos\alpha\sin^2\alpha\\ \Rightarrow a^2 = \cos^2\alpha -8\cos^2\alpha\sin^2\alpha +16\cos^2\alpha\sin^4\alpha =\\ = \cos^2\alpha + 8\cos^2\alpha\sin^2\alpha(2\sin^2 \alpha - 1)$$
And:
$$b = 3\sin\alpha -4\sin^3\alpha = 3\sin\alpha - 4\sin\alpha(1-\cos^2\alpha) = -\sin\alpha + 4\sin\alpha\cos^2\alpha\sin\alpha \\ b^2 = \sin^2\alpha - 8\sin^2\alpha\cos^2\alpha + 16\sin^2\alpha\cos^4\alpha =\\ = \sin^2\alpha + 8\sin^2\alpha\cos^2\alpha(2\cos^2\alpha - 1)$$
Therefore:
$$a^2 + b^2 = \cos^2\alpha + \sin^2\alpha + 8\sin^2\alpha\cos^2\alpha(2\sin^2\alpha + 2\cos^2\alpha - 2) = 1$$
A: 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 $\theta=3\alpha$).
A: Hint: in fact, $a = \cos 3\alpha$ and $b = \sin 3 \alpha$
A: To solve it using the approach the OP wants to take, it helps to use the abbreviations $c$ and $s$ for $\cos\alpha$ and $\sin\alpha$.  With these abbreviations, the desired identity becomes
$$16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)=1$$
The key thing we know is that $c^2+s^2=1$.  Squaring this gives
$$1=(c^2+s^2)^2=c^4+2c^2s^2+s^4\implies c^4+s^4=1-2c^2s^2$$
Cubing it gives
$$1=(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6\implies c^6+s^6=1-3c^2s^2(c^2+s^2)=1-3c^2s^2$$
And now we're ready to plug these in:
$$\begin{align}
16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)
&=16(1-3c^2s^2)-24(1-2c^2s^2)+9\\
&=16-24+9-48c^2s^2+48c^2s^2\\
&=1
\end{align}$$
