De Moivre's formula to solve $\cos$ equation Use De Moivre’s formula to show that
$$
\cos\left(x\right) + \cos\left(3x\right) = 2\cos\left(x\right)\cos\left(2x\right)
$$
$$
\mbox{Show also that}\quad
\cos^{5}\left(x\right) = \frac{1}{16}\,\cos\left(5x\right) + \frac{5}{16}\,\cos\left(3x\right) + \frac{5}{8}\,\cos\left(x\right)
$$
Hence solve the equation $\cos\left(5x\right) = 16\cos^{5}\left(x\right)$ completely. 
Express your answers as rational multiples of $\pi$.
 A: Let me first show you another problem using DeMoivre's formula, to give you the idea of how to do the first two parts. Namely, I will show that $$\cos(4x)=8\cos^4x-8\cos^2x+1,$$ using only DeMoivre's formula and the Pythagorean identity $\sin^2x=1-\cos^2x$.
First off, we know that $$\cos(4x)+i\sin(4x)=(\cos x+i\sin x)^4,$$ and by expanding the right-hand side, we have $$\cos(4x)+i\sin(4x)=\cos^4x+4i\cos^3x\sin x-6\cos^2x\sin^2x-4i\cos x\sin^3x+\sin^4x.$$ Gathering our real and imaginary terms gives us $$\cos(4x)+i\sin(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x+i(4\cos^3x\sin x-4\cos x\sin^3x),$$ so, remembering that sine and cosine are real-valued functions on the reals, we have that $$\cos(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x.$$ At that point, we can rewrite as $$\cos(4x)=\cos^4x-6\cos^2x\sin^2x+(\sin^2x)^2,$$ which means that $$\cos(4x)=\cos^4x-6\cos^2x(1-\cos^2x)+(1-\cos^2x)^2,$$ whence expanding and collecting like terms gives us $$\cos(4x)=8\cos^4x-8\cos^2x+1,$$ as desired.

Now, once we've proved that $$\cos(x)+\cos(3x)=2\cos(x)\cos(2x)\tag{$\heartsuit$}$$ and that $$\cos^5(x) = \frac1{16}\cos(5x) + \frac5{16}\cos(3x) + \frac58\cos(x),\tag{$\diamondsuit$}$$ let us assume that $$\cos(5x)=16\cos^5(x).\tag{$\spadesuit$}$$ By $(\spadesuit)$ and $(\diamondsuit),$ we can conclude that $$0=\cos(3x)+2\cos(x).$$ (Do you see how?) By $(\heartsuit),$ it then follows that $$0=2\cos(x)\cos(2x)+\cos(x)\\0=\cos(x)\bigl(2\cos(2x)+1\bigr).$$ (Do you see how?) Thus, we have $$\cos(x)=0$$ or $$\cos(2x)=-\frac12,$$ which I leave to you to solve.
A: HINT:
$$ (\cos x + i \sin x)^n = \cos nx + i \sin nx $$
A: @Lauren, the approach comes down to the fact that $Re[(\text{cos}(x)+i\cdot \text{sin}(x))^3] = \text{cos}(3x)$.
So, $$\text{cos}(3x) \\ = \text{cos}^3(x) + 3\text{cos}(x)\cdot (i^2\cdot \text{sin}^2(x)) \\ = \text{cos}(x)\cdot (\text{cos}^2(x) - 3\text{sin}^2(x)) \\ = \text{cos}(x)\cdot (1-2\text{sin}^2(x)+ \text{cos}(2x)-1) \\ = \text{cos}(x)\cdot (2\text{cos}(2x) - 1) = \text{cos}(3x).$$
This proves part 1. Then the next part follows from the fact that $\text{cos}(3x) = -2\text{cos}(x)$. Why? I'll leave this as an exercise.
