To prove the double-angle identites, we just have to use the identities proven in addition and subtraction identities, and let \[\alpha=\beta=\theta\]
\begin{align*}
\sin(\alpha+\beta)&=\sin\alpha\cos\beta+\cos\alpha\sin\beta\\
\sin(\theta+\theta)&=\sin\theta\cos\theta+\cos\theta\sin\theta\\
\therefore \sin(2\theta)&=2\sin\theta\cos\theta
\end{align*}
\begin{align*}
\cos(\alpha+\beta)&=\cos\alpha\cos\beta-\sin\alpha\sin\beta\\
\cos(\theta+\theta)&=\cos\theta\cos\theta-\sin\theta\sin\theta\\
\therefore \cos(2\theta)&=\cos^{2}\theta-\sin^{2}\theta\\
\end{align*}
Since \[\sin^{2}\theta+\cos^{2}\theta=1\],
\begin{align*}
\cos(2\theta)&=(1-\sin^{2}\theta)-\sin^{2}\theta\\
\therefore\cos(2\theta)&=1-2\sin^{2}\theta\\
\end{align*}
\begin{align*}
\cos(2\theta)&=\cos^{2}\theta-(1-\cos^{2}\theta)\\
\therefore\cos(2\theta)&=2\cos^{2}\theta-1
\end{align*}
Similarly,
\begin{align*}
\tan(\theta+\theta)&=\frac{\tan\theta+\tan\theta}{1-\tan\theta\tan\theta}\\
\therefore \tan(2\theta)&=\frac{2\tan\theta}{1-\tan^{2}\theta}
\end{align*}