How to Derive a Double Angle Identity. How does one derive the following two identities:
$$\begin{align*}
\cos 2\theta &= 1-2\sin^2\theta\\
\sin 2\theta &= 2\sin\theta\cos\theta
\end{align*}$$

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 $\sin2\theta$.)
