<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] a) Since $\sin^2(\theta) + \cos^2(\theta) =1 $ we can write: $1=1^2= (\sin^2(\theta) + \cos^2(\theta))^2= \cos^4(\theta)+\sin^4(\theta)+2 \sin^2(\theta) \cos^2(\theta)$. Hence $\cos^4(\theta)+\sin^4(\theta) = 1-2\sin^2(\theta)\cos^2(\theta) = 1 - \frac{1}{2} \sin^2(2\theta) = 1 - \frac{1}{2} \frac{1-\cos(4\theta)}{2} $, the claim follows. b) is similar, even there you just have to remember basic goniometric ids (in particular bisection and dupliaction).
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