How to prove the following equation by a quick method?
\begin{eqnarray}
\\\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)\\
\end{eqnarray}
If I use so much time to expand it and take extra care of the calculation process, I can find the answer. Since it is so easy to get mistakes, I have to try it 3 time to obtain the correct answer. Therefore, I wonder there are other easy ways to deal with this question. Is it right?
Thank you for your attention
Best Answer
Using $a^3+b^3=(a+b)^3-3ab(a+b),$
$$(\sin^2x)^3+(\cos^2x)^3=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$$
$$=1-3\frac{(2\sin x\cos x)^2}4$$
$$=1-\frac34(\sin^22x)\text{ using }\sin2x=2\sin x\cos x$$
$$=1-\frac34\frac{(1-\cos4x)}2 \text{ using }\cos2y=1-2\sin^2y$$