How is $\cos2\theta = \cos^2\theta- \sin^2\theta$? 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)
 A: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, take $a=b$
A: I think it is worth demonstrating the validity of the formula
\begin{align*}
\cos(\theta + \varphi) = \cos(\theta)\cos(\varphi) - \sin(\theta)\sin(\varphi)
\end{align*}
To begin with, I would recommend to start from the fact that the composition of two rotations $\theta$ and $\varphi$ is given by a rotation of $\theta + \varphi$. Therefore let us consider the standard basis $\mathcal{B} = \{(1,0),(0,1)\}$ of $\textbf{R}^{2}$.
Given that $(x,y)_{\mathcal{B}}\in\textbf{R}^{2}$, its new coordinates (on the same basis) after a rotation of angle $\alpha$ is given by $(x',y')_{\mathcal{B}} = (\cos(\alpha)x - \sin(\alpha)y,\sin(\alpha)x + y\cos(\alpha))_{\mathcal{B}} = T_{\alpha}(x,y)$ (draw it if you need), we have that
\begin{align*}
[T_{\alpha}]_{\mathcal{B}} = 
\begin{bmatrix}
\cos(\alpha) & -\sin(\alpha)\\
\sin(\alpha) & \cos(\alpha)
\end{bmatrix}
\end{align*}
Consequently, one has that
\begin{align*}
[T_{\theta+\varphi}]_{\mathcal{B}} & =
\begin{bmatrix}
\cos(\theta + \varphi) & -\sin(\theta + \varphi)\\
\sin(\theta + \varphi) & \cos(\theta + \varphi) 
\end{bmatrix}\\\\
& =
\begin{bmatrix}
\cos(\theta) & -\sin(\theta)\\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
\cos(\varphi) & -\sin(\varphi)\\
\sin(\varphi) & \cos(\varphi)
\end{bmatrix}\\\\
& = \begin{bmatrix}
\cos(\varphi)\cos(\varphi) - \sin(\theta)\sin(\varphi) & -\cos(\theta)\sin(\varphi) - \cos(\varphi)\sin(\theta)\\
\sin(\theta)\cos(\varphi) + \cos(\theta)\sin(\varphi) & -\sin(\theta)\sin(\varphi) + \cos(\theta)\sin(\varphi)
\end{bmatrix} = [T_{\theta}]_{\mathcal{B}}[T_{\varphi}]_{\mathcal{B}}
\end{align*}
from whence we conclude that
\begin{align*}
\cos(\theta + \varphi) = \cos(\theta)\cos(\varphi) - \sin(\theta)\sin(\varphi)
\end{align*}
holds indeed. Consequently, one has that
\begin{align*}
\cos(2\theta) = \cos(\theta)\cos(\theta) - \sin(\theta)\sin(\theta) = \cos^{2}(\theta) - \sin^{2}(\theta)
\end{align*}
just as desired. Hopefully it helps.
A: 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
$$\begin{cases}\cos(2x)=\dfrac{AB}{CD}=\dfrac{\cos(x)}{CD}\\BC^2=(CD+\sin(x))^2+\cos^2(x)\\\dfrac{BC}{CD}=\dfrac{BA}{AD}\iff BC=\dfrac{\cos(x)}{\sin(x)}\cdot CD\end{cases}$$ 

A: $$\cos2\theta+i\sin2\theta=e^{2i\theta}=(e^{i\theta})^2=(\cos\theta+i\sin\theta)^2=\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta.$$
Two formulas for the price of one !
