How to Derive a Double Angle Identity. How does one derive the following two identities:
$$\begin{align*}
\cos 2\theta &= 1-2\sin^2\theta\\
\sin 2\theta &= 2\sin\theta\cos\theta
\end{align*}$$
 A: Hints:
For the $\cos 2\theta$ formula, use the sum identity (with $x=y=\theta$)
$$
\cos(x+y)=\cos x\cos y - \sin x\sin y,
$$
followed by the Pythagorean identity $\cos^2 x=1-\sin^2 x $.

For the $\sin 2\theta$ formula , use the sum identity (with $x=y=\theta$)
$$
\sin(x+y)=\sin x\cos y +\sin y\cos x.
$$ 



Or, for the $\cos2\theta$ formula and $0<\theta<\pi/2$, consider the diagram:
 
We have
$$
\cos\theta ={ {1+\cos2\theta}\over\sqrt{2+2\cos 2\theta}};
$$
whence
$$
\sqrt 2\cos\theta=\sqrt{1+1\cos\theta},
$$
or,
$$
2\cos^2\theta= 1+1\cos2\theta
$$
From this, we have
$$
2-2\sin^2\theta=1+\cos2\theta,
$$
or
$$
\cos2\theta =1-2\sin^2\theta.
$$
(Having this in hand, we could also use the diagram to derive the formula for $\sin2\theta$.)
A: $e^{i\theta}$ means the point with angle $\theta$ on the unit circle. Euler formula tells us that $e^{i\theta}=\cos \theta+i\sin \theta$. Then 
$e^{i2\theta}=\cos^2 \theta-\sin^2 \theta+2i\sin \theta\cos\theta$. Also, we know $e^{i(2\theta)}=\cos 2\theta+i\sin 2\theta$. Compare the real and imaginary parts, and you get the desired result.
All I assume is that you know Euler formula.             
A: The first one follows from:
$$\cos 2\theta = \cos(\theta +\theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta .$$ Now use the fact $\cos^2 \theta + \sin^2 \theta =1.$  
The second one follows from 
$$ \sin 2\theta = \sin(\theta +\theta) = \sin \theta \cos \theta + \cos \theta \sin \theta. $$
