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.

Hint: For $a$
$$L.H.S=\sin^4 A+ \cos^4 A$$
$$=(\sin^2 A+ \cos^2 A)^2-2\sin^2A.\cos^2A$$
$$=1-\frac {1}{4}(2\sin A.\cos A)^2$$
$$=1-\frac {1}{4} \sin^2 2A$$
$$=\frac {4-\sin^2 2A}{4}$$
$$=\frac {4-1+\cos 4A}{4}$$
$$=\frac {3+\cos 4A}{4}=R.H.S$$ 
Proved.
Just consider that $\theta=A$.
Follow similar steps above for $b$.
