How do I prove that $\cos\left(2x\right)=1-2\sin^2\left(x\right)$? While trying to solve the equation $\sin\left(x\right)=\cos\left(2x\right)$, a user on this forum suggested that I turn the equation into a quadratic form by converting $\cos(2x)$ using the identity $\cos\left(2x\right)=1-2\sin^2\left(x\right)$.
What is the logic behind this identity and how can I derive it?
 A: This identity follows from the more general
$$\cos (x+y)=\cos x \cos y - \sin x \sin y$$
by $y=x$, that is
$$\cos (2x)=\cos (x+x)=\cos x \cos x - \sin x \sin x=\cos^2 x-\sin^2 x=1-2\sin^2 x$$
using that $\cos^2 x=1-\sin^2 x$.
Refer also to the related


*

*Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$

*How can I understand and prove the "sum and difference formulas" in trigonometry?
A: $$
\\\cos(2x)=\frac{(e^{2ix}+e^{-2ix})}{2}=1-2(\frac{e^{ix}-e^{-ix}}{2i})^2=1-2\sin^2{x}
$$
A: Well, you'd first have to know the following identity.
$$\cos(a\pm b) = \cos a\cos b\mp \sin a\sin b$$
The diagram below demonstrates it perfectly.

From here, apply the general identity.
$$\cos(2a) = \cos(a+a) = \cos a\cos a-\sin a\sin a$$
$$\implies \color{blue}{\cos(2a) = \cos^2 a-\sin^2 a}$$
This can be rewritten by using $\color{purple}{\cos^2 a = 1-\sin^2 a}$ or $\color{purple}{\sin^2 a = 1-\cos^2 a}$.
$$\implies \cos(2a) = \cos^2 a-\color{purple}{(1-\cos^2 a)} = 2\cos^2 a-1$$
$$\implies \cos(2a) = \color{purple}{1-\sin^2 a}-\sin^2 a = 1-2\sin^2 a$$
Therefore, all three expressions are valid for $\cos(2a)$.
Edit: I am now aware that the diagram above is by Blue. You may want to check this link as well: 
How can I understand and prove the "sum and difference formulas" in trigonometry?
A: Define $f(t)=\cos(2x-t)\cos(t)-\sin(2x-t)\sin(t)$.  We have $f'(t)=0$, hence $f$ is constant.  From $f(0)=f(x)$ follows $\cos(2x)=\cos^2(x)-\sin^2(x)$.  Now use Pythagoras.
See also Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$, please.
A: write $$\cos (2x)=\cos(x+x)$$
you know $$\cos(x+y)=\cos x\cdot \cos y-\sin x\cdot \sin y$$
So, $$\cos(2x)=\cos x\cdot \cos x-\sin x\cdot \sin x$$
Or, $$\cos(2x)=\cos^2x-\sin^2x$$
write $\cos^2x$ as $1-\sin^2x$ 
So, $$\cos(2x)=1-\sin^2x-\sin^2x$$

you get$$\cos(2x)=1-2\sin^2x$$ 

Similarly get $$\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1$$
A: $$\cos(2x)=\cos(x+x)=\cos(x)\cos(x)-\sin(x)\sin(x)=\cos^2(x)-\sin^2(x)$$
then we know that $cos^2(x)+\sin^2(x)=1$ so:
$$cos(2x)=1-2\sin^2(x)$$
you can find proof of the compound angle formula here:
https://www.math-only-math.com/proof-of-compound-angle-formula-sin-alpha-plus-beta.html
