Power Reduction Formula

Q1: How can I simplify $cos^2(x)$?
A1: Recall that Euler's formula implies that:
$cos(x)=\frac{1}{2}(e^{jx}+e^{-jx})$

Hence,
$cos^2(x)=\frac{1}{4}(e^{jx}+e^{-jx})^2$
$cos^2(x)=\frac{1}{4}(e^{2jx}+2+e^{-2jx})$
$cos^2(x)=\frac{1}{2}(\frac{e^{2jx}+e^{-2jx}}{2}+1)$
$cos^2(x)=\frac{1}{2}(cos(2x)+1)$
$cos^2(x)=\frac{1+cos(2x)}{2}$

Q2: How can I simplify $sin^2(x)$?
A2: Using similar logic as in A1:
$sin(x)=\frac{1}{2j}(e^{jx}-e^{-jx})$

Hence,

$sin^2(x)=-\frac{1}{4}(e^{jx}-e^{-jx})^2$
$sin^2(x)=-\frac{1}{4}(e^{2jx}-2+e^{-2jx})$
$sin^2(x)=-\frac{1}{2}(\frac{e^{2jx}+e^{-2jx}}{2}-1)$
$sin^2(x)=-\frac{1}{2}(cos(2x)-1)$
$sin^2(x)=\frac{1-cos(2x)}{2}$

Q3: How can I simplify $tan^2(x)$?
A3: Using the results of A1 and A2:
$tan(x)=\frac{sin(x)}{cos(x)}$
$tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$
$tan^2(x)=\frac{1-cos(2x)}{1+cos(2x)}$