Why $\cos^3 x - 2 \cos (x) \sin^2(x) = {1\over4}(\cos(x) + 3\cos(3x))$? Wolfram Alpha says so, but step-by-step shown skips that step, and I couldn't find the relation that was used.
 A: Here's a detailed step by step:
$=\frac{1}{4}(\cos(x)+3\cos(3x))$
$=\frac{1}{4}(\cos(x)+3\cos(x+2x))$
$=\frac{1}{4}(\cos(x)+3[\cos(x)\cos(2x)-\sin(x)\sin(2x)])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)(1-2\sin^2(x))-\sin(x)(2\sin(x)\cos(x))])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)-2\cos(x)\sin^2(x))-2\sin^2(x)\cos(x)])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)-4\cos(x)\sin^2(x))])$
$=\frac{1}{4}(\cos(x)+3\cos(x)-12\cos(x)\sin^2(x))$
$=\frac{1}{4}(4\cos(x)-12\cos(x)\sin^2(x))$
$=\cos(x)-3\cos(x)\sin^2(x))$
$=\cos(x)-\cos(x)\sin^2(x)-2\cos(x)\sin^2(x)$
$=\cos(x)(1-\sin^2(x))-2\cos(x)\sin^2(x)$
$=\cos^3(x)-2\cos(x)\sin^2(x)$
If your question is an assignment, please try to resolve it from scratch without looking at the answer. There are other ways to solve it. Also try that for practice.
A: You can prove the above relation by using derivatives. i.e., prove that LHS and RHS have the same derivatives. So LHS-RHS=c for some constant c. Now show that c must be zero by setting x=0.
A: Hint: Taking the real part of De Moivre's formula
$$
\cos(3x)+i\sin(3x)=(\cos(x)+i\sin(x))^3
$$
and applying $\cos^2(x)+\sin^2(x)=1$, we have
$$
\begin{align}
\cos(3x)
&=\cos^3(x)-3\sin^2(x)\cos(x)\\
&=4\cos^3(x)-3\cos(x)
\end{align}
$$
Furthermore, the left side is
$$
\cos^3(x)-2\cos(x)\sin^2(x)=3\cos^3(x)-2\cos(x)
$$
A: Start by the left hand side. You can calculate it known that:


*

*$\cos (x+2x)=\cos x\cos 2x-\sin x\sin 2x,$

*$\cos 2x=2\cos^2 x -1,$

*$\sin 2x=2\sin x\cos x,$

*$\cos^2 x+\sin^2 x=1.$
$$\frac{1}{4}\left(\cos x+3\cos 3x\right)=\frac{1}{4}\left(\cos x+3\left(4\cos^3 x-3\cos x\right)\right)=3\cos^3 x-2\cos x=\cos^3 x-2\cos x\left(1-\cos^2 x\right)$$
