Why $\cos^3 x - 2 \cos (x) \sin^2(x) = {1\over4}(\cos(x) + 3\cos(3x))$?

Question

Wolfram Alpha says so, but step-by-step shown skips that step, and I couldn't find the relation that was used.

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ \cos^3(x)-2\cos(x)\sin^2(x)=3\cos^3(x)-2\cos(x) $$

Answer 4

Start by the left hand side. You can calculate it known that:

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

• $\cos 2x=2\cos^2 x -1,$

• $\sin 2x=2\sin x\cos x,$

• $\cos^2 x+\sin^2 x=1.$

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