How is $\cos2\theta = \cos^2\theta- \sin^2\theta$?

Question

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)

Answer 1

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, take $a=b$

Answer 2

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.

Answer 3

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

$$\begin{cases}\cos(2x)=\dfrac{AB}{CD}=\dfrac{\cos(x)}{CD}\\BC^2=(CD+\sin(x))^2+\cos^2(x)\\\dfrac{BC}{CD}=\dfrac{BA}{AD}\iff BC=\dfrac{\cos(x)}{\sin(x)}\cdot CD\end{cases}$$

Answer 4

$$\cos2\theta+i\sin2\theta=e^{2i\theta}=(e^{i\theta})^2=(\cos\theta+i\sin\theta)^2=\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta.$$

Two formulas for the price of one !