<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] There is a general systematic method to prove this kind of linearisation identities. It's not as swift as the previous answers, bit it's purely automatic.
You use the Euler formulae
$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}2 \qquad \text{et}\qquad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
and you expand the powers to get an expression with terms in $\cos(p\theta)$ and $\sin(q\theta)$.
\begin{align*}
\cos^4(\theta) &= \left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^4\\
&= \frac{e^{i4\theta} + 4 e^{i2\theta} + 6 + 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 + \frac{\cos 2\theta}2 + \frac 38.\\
\sin^4(\theta) &= \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^4\\
&= \frac{e^{i4\theta} - 4 e^{i2\theta} + 6 - 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 - \frac{\cos 2\theta}2 + \frac 38.\\
\text{therefore}\qquad \cos^4\theta+\sin^4\theta &= \frac{\cos 4\theta}4 + \frac 34.
\end{align*}
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