Proof that $\cos(2\theta) = \cos^2\theta-\sin^2\theta$ by showing equivalence of derivatives It can be shown using trig identities that $\cos(2\theta) = \cos^2\theta-\sin^2\theta$. 
But if we let $f(x) = \sin(2x)$, we can differentiate two ways:
1) $$f(x) = \sin(2x) \rightarrow f(x) = 2\sin(x)\cos(x)$$
Differentiating using the product rule we see that :
$$f^{'}(x) = 2[\cos(x)\cos(x)-\sin(x)\sin(x)] = 2[\cos^2(x)-\sin^2(x)]$$
2) If we differentiate $f(x)$ as is, then :
$$\frac{d}{dx}f(x) = 2\cos(2x) $$
Therefore: $2\cos(2x) = 2[\cos^2(x) - \sin^2(x)] \rightarrow \cos(2x) = \cos^2(x) -\sin^2(x)$
Is this a viable proof?
Thanks in Advance!
 A: Sure, totally valid. There are other interesting ways like this that also work for verifying other trig identities. I recall proving one through Fourier series back in August for a class, an example.
A bit overkill, obviously, but 100% valid, as with your method.
A: The ideas and work shown are valid.
In order to format the work here into the format of a proof, we need to identify our hypotheses.
The things that are assumed here are that $\sin{2x}=2\sin{x}\cos{x},$ (I am calling this an assumption, since the double angle formula for cosine is not implicitly assumed), the derivative of sin is cos, and the derivative of cos is -sin.
Usually, the double angle formula for cos is used to prove that the derivatives of the trigonometric functions are what they are. So explicitly calling those things assumptions allow this to not be circular logic.
A: The nethod is valid and its logic is:


*

*$f(x)$ has a unique derivative, $f'(x)$, which can be found two ways.

*One way gives $f'(x)=g(x)$, and the other gives $f'(x)=h(x)$.

*Since $f'(x)$ is unique, $g(x)=h(x)$.


It would only be invalid if, say, one of the differentiation methods was wrong.
