$\sin 2\alpha=2\sin \alpha\cos \alpha$
$\cos 2\alpha=2\cos^{2} \alpha-1=1-2\sin^{2}\alpha$
$\tan 2\alpha=\frac{2\tan \alpha}{1-\tan^{2} \alpha}$
Proof by Addition Formulas
If we Set $\beta=\alpha$ in Addition Formulas ： $ \sin (\alpha + \beta)=\sin \alpha\cos \beta+ \cos \alpha\sin \beta$, we obtain
$ \sin 2\alpha=\sin (\alpha + \alpha)=\sin \alpha\cos \alpha+ \cos \alpha\sin \alpha=2\sin \alpha\cos \alpha$
Others can be proved similarly.
Proof by De Moivre’s Theorem
By De Moivre’s Theorem, we have
$$(\cos \theta+i\sin \theta)^{2}=\cos 2\theta+i\sin \theta$$
Then if we expand LHS of the above equation
$$\begin{eqnarray*}(\cos \theta+i\sin \theta)^{2}&=&\cos^{2}\theta +2i\sin\cos-\sin^{2}\theta\\&=&( \cos^{2}\theta-\sin^{2}\theta ) + i ( 2\sin\cos )\end{eqnarray*}$$
Now, if we compare the real part and the imaginary part on the left and the right side respectively, we obtain
$$\cos 2\theta=\cos^{2} \theta-\sin^{2}\theta$$
$$\sin 2\theta=2\sin \theta\cos \theta$$