<s>[INST] 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.
 [/INST] 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:
$$a^2 + b^2 = \cos^2\alpha + \sin^2\alpha + 8\sin^2\alpha\cos^2\alpha(2\sin^2\alpha + 2\cos^2\alpha - 2) = 1$$
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