Using Complex exponential definitions of sine and cosine, prove $\cos2\theta=\cos^2\theta-\sin^2\theta$ Question

Please do not just tell me the answer, please provide helpful hints and hide the answers 

Using Complex exponential definitions of sine and cosine, prove $\cos\theta=\cos^2 \theta-\sin^2\theta$

All that I know is the trig identity:

$\cos2\theta=1-2\sin^2\theta$ or $\cos2\theta=2\cos^2\theta-1$
Hint:


  $\cos 2\theta + i \sin 2\theta = e^{2i\theta} =\left(e^{i\theta}\right)^2=(\cos \theta+ i\sin \theta)^2$


Then expand the right side and compare real parts
$\begin{array}{lcl}\cos(2\theta)+i\sin(2\theta) &amp; = &amp; e^{2i\theta} \\ &amp; = &amp; (e^{i \theta})^2 \\ &amp; = &amp;  (\cos\theta+i\sin\theta)^2 \\ &amp; = &amp; (\cos\theta)^2+2i\cos θ\sin θ+i^2(\sin θ)^2 \\ &amp; = &amp; (\cos θ)^2+2i\cos θ\sin θ−(\sin θ)^2\end{array}$

Equating real and imaginary parts gives us:

$\begin{array}{lcl}\cos(2θ)  =  (\cosθ)^2−(\sinθ)^2 \\ \sin(2θ)  = 2\cosθ\sinθ \end{array}$