TITLE: How is $\cos2\theta = \cos^2\theta- \sin^2\theta$?
QUESTION [5 upvotes]: On Khan Academy in this video at 3:53 minutes in, Khan uses the equation $\cos2\theta = \cos^2\theta- \sin^2\theta$, where can I view the proof for this statement? (He doesn't explain this part in that particular video)

REPLY [2 votes]: HINT.- I give you a nice proof using a little of algebra and geometry with the angle bisector theorem. In the attached figure you have $BC$ and $CD$ to eliminate in the system
$$\begin{cases}\cos(2x)=\dfrac{AB}{CD}=\dfrac{\cos(x)}{CD}\\BC^2=(CD+\sin(x))^2+\cos^2(x)\\\dfrac{BC}{CD}=\dfrac{BA}{AD}\iff BC=\dfrac{\cos(x)}{\sin(x)}\cdot CD\end{cases}$$