Knowing that $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ show that $\cos(2x) \equiv \cos^2x - \sin^2x $ Apparently one option is to differentiate the identity $ \sin(2x) \equiv 2 \sin(x)\cos(x) $ to get the identity $\cos(2x) \equiv \cos^2x - \sin^2x $. Which is surprising as I didn't realize that differentiating an identity produces another identity.
However, I'd like to know how it can be done without involving calculus.
I have the list of all trigonometric properties and operations at hand, but can't find the right way to relate them to go from one to the other.
Could I get a hint?
 A: $$\cos^2 (2x) = 1 - \sin^2 (2x) = 1 - 4 \sin^2 x \cos^2 x = 1 - 4 \sin^2 x (1 - \sin^2 x)$$
$$ = 4 \sin^4 x - 4 \sin^2 x +1$$
Now if we let $u = \sin x$, we have $4u^4 - 4u^2 + 1 = (2u^2-1)^2$. Thus:
$$\cos^2 (2x) = (2 \sin^2 x - 1)^2$$
$$\cos^2 (2x) = (2 \sin^2 x - (\sin^2 x +\cos^2 x) )^2$$
and now you are very close to the desired result.
One last note: from $a^2 = b^2$ we cannot directly conclude that $a = b$. However, we can disprove that $\cos(2x) = (\sin^2 x - \cos^2 x)$ by substituting in $x = 0$ for example: this gives $1 = -1$ which is false. Since what we have done is true no matter which $x$ we choose, the other possibility $\cos(2x) = -(\sin^2 x - \cos^2 x) = \cos^2 x - \sin^2 x$  must be true.
A: $$\cos^22x=1-\sin^22x=1-4\sin^2x\cos^2x=(\cos^2x+\sin^2x)^2-4\sin^2x\cos^2x
\\=(\cos^2x-\sin^2x)^2.$$

We still have to check the sign. $\cos2x$ is positive in the first and fourth quadrants, and this corresponds to ranges of $x$ such that $|\cos x|\ge|\sin x|$, and the original claim is valid for all $x$.
A: You can also make a substitution $\displaystyle x\to\left( \frac{π}{4}-t\right)$. You get $$\begin{align} \sin\left(\frac{π}{2}-2t\right)&=2\sin\left(\frac{π}{4}-t\right)\cos\left(\frac{π}{4}-t\right)\\ &=2\left(\frac{1}{\sqrt 2}\cos t-\frac{1}{\sqrt 2}\sin t\right)\left(\frac{1}{\sqrt 2}\cos t+\frac{1}{\sqrt 2}\sin t\right)\\ \cos 2t&=\cos^2 t-\sin^2 t\end{align}$$
