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

$$\begin{array}{rcl}(\cos \theta +i\sin \theta {)}^2& =& {\cos }^2\theta +2i\sin \cos -{\sin }^2\theta \\ & =& ({\cos }^2\theta -{\sin }^2\theta )+i(2\sin \cos )\end{array}$$

Now, if we compare the real part and the imaginary part on the left and the right side respectively, we obtain

$$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$$

$$\sin 2\theta =2\sin \theta \cos \theta$$