What proves this trigonometric identity $\cos2 \theta=\cos^2\theta-\sin^2\theta$?

Question

I am trying to understand why $$\cos2 \theta=\cos^2\theta-\sin^2\theta$$

I cannot find any information on this identity on the Wikipedia page listing the trigonometric identities

Answer 1

It comes from the more general formula $$ \cos ( x+y) = \cos (x) \cos (y) - \sin (x) \sin(y)$$

If you let x=y, you will get $$ \cos ( x+x) = \cos (x) \cos (x) - \sin (x) \sin(x)$$

Which is the same as $$\cos2 x=\cos^2 x-\sin^2 x$$

Answer 2

The linked wikipedia page gives the standard angle-addition formula for cosine,

$\cos( \alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta, \tag 1$

and provides a demonstration that it is so; given (1), we have

$\cos 2 \theta = \cos (\theta + \theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta; \tag 2$

so that's one way to arrive at the desired formula.

I prefer to us the matrix exponential:

$e^{i \psi} = \cos \psi + i \sin \psi, \; \psi \in \Bbb R; \tag 2$

then

$\cos(\psi + \phi) + i \sin (\psi + \phi) = e^{i(\psi + \phi)} = e^{i\psi}e^{i\phi}$ $= (\cos \psi + i \sin \psi)(\cos \phi + i \sin \phi)$ $= \cos \psi \cos \phi - \sin \psi \sin \phi + i(\sin \psi \cos \phi + \sin \phi \cos \psi); \tag 3$

equating the real parts on either side of (3) yields

$\cos(\psi + \phi) = \cos \psi \cos \phi - \sin \psi \sin \phi, \tag 4$

essentially the same as (1), easily seen to lead to (2) by taking $\theta = \psi = \phi$.