<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] 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:
$$\begin{align}
16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)
&=16(1-3c^2s^2)-24(1-2c^2s^2)+9\\
&=16-24+9-48c^2s^2+48c^2s^2\\
&=1
\end{align}$$
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