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.

$\cos(m\theta)$ is a polynomial with degree $m$ in $\cos\theta$, in particular
$$ \cos(4\theta) = T_4(\cos\theta) = 8\cos^4\theta-8\cos^2\theta + 1\tag{1}$$
so the given identities are equivalent to the polynomial identities
$$ (1-x^2)^2+x^4 = \frac{3+T_4(x)}{4},\qquad (1-x^2)^3+x^6 = \frac{5+3 T_4(x)}{8}\tag{2}$$
that directly follow from $(1)$.
