Double Angle Formula

Given

$$e^{+\imath \theta} = \cos \theta + \imath \sin \theta\ \ \;\ \ e^{+\imath \theta} = \cos \theta + \imath \sin \theta$$

the product of these two is

$$e^{+\imath \theta} e^{+\imath \theta} = (\cos \theta + \ima... ...cos^2 \theta - \sin^2 \theta +\imath(2 \sin\theta \cos \theta)$$

But

$$e^{+\imath \theta} e^{+\imath \theta} = e^{\imath (\theta+ \... ...cos^2 \theta - \sin^2 \theta +\imath(2 \sin\theta \cos \theta)$$

which implies two equalities, one for the real part, and one for the imaginary part of these two equalities:

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

and

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

.