<s>[INST] Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
 [/INST] Hint:
$$\sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
$$\cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
</s>