How can I prove the following equation: $$\cos^2(x) = \frac 12\big(1+\cos(2x)\big),\tag1$$ using Euler's identity? $$e^{i\pi} + 1 = 0.\tag*{$\begin{align} \because e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \text{cis} \ \theta \end{align}$}$$
I have tried equating Euler's equation to cos on one side and squaring that but haven't had luck reducing it to the desired form as outlined in $(1)$.
On
An alternative approach (without Euler's Identity).
Recall that $\sin(2x) = 2\cos(x)\sin(x)$. Negate both sides and integrate:$$\frac{1}{2}\cos(2x) + C = \cos^2(x)$$
Plugging in $x = 0$ yields:
$$\frac{1}{2}\cos(0) + C = \cos^2(0)$$
and, noting that $\cos(0) = 1$, this latter equation becomes:
$$\frac{1}{2} + C = 1$$
from which we conclude that $C = 1/2$. Plugging this value of $C$ into the first centered equation above and factoring out a $1/2$ leaves us with the desired identity:
$$\frac{1}{2}(\cos(2x) + 1) = \cos^2(x)$$
$$\begin{align}e^{ix}&=\cos x+i\sin x\\e^{-ix}&=\cos x-i\sin x\\\hline\color{red}{\cos x}&=\color{blue}{\dfrac12(e^{ix}+e^{-ix})}\\\cos^2x&=\dfrac14(e^{2ix}+2+e^{-2ix})\\&=\dfrac12\cdot\dfrac12\left(e^{2ix}+e^{-2ix}+2\right)\\&=\dfrac12\left\{1+\dfrac12(e^{2ix}+e^{-2ix})\right\}\\\color{red}{\cos^2x}&=\color{blue}{\dfrac12(1+\cos 2x)}\end{align}$$