How I can prove that the derivative of $\sin x$ is $\cos x$? I want to prove that the derivative of $\sin x$ is $\cos x$, but how? Is this proposition true?

$$(\sin x) ^{\prime}=\cos x \Longleftrightarrow‎ \sin x ^{2}+\cos x ^{2}=1 , \forall x$$
Thank you.
$$f(x)=\sin x$$
Using first principle
$$f'(x)=\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$
$$=\lim_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)\cos\left(\frac{2x-h}{2}\right)}{\frac{h}{2}}$$
We know that $$ f'(x) = \frac{d}{dx} f(x) = \lim_{h\to 0}\frac{f(x+h) -f(x)}{h}$$
Take $$f(x)=\sin x \Leftrightarrow f(x+h) = \sin(x+h) $$
Substitute them in the fundamental relation of derivative of function with limits.
$$\frac{d}{dx} \sin x =\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$
$$=\lim_{h\to 0}\frac{2\sin \Big(\frac{h}{2}\Big)\cos\Big(\frac{2x-h}{2}\Big)}{h}$$
$$=\lim_{h\to 0}\frac{\sin \Big(\frac{h}{2}\Big)\cos\Big(\frac{2x-h}{2}\Big)}{\frac{h}{2}}$$
$$=\lim_{\frac{h}{2}\to 0}\frac{\sin \Big(\frac{h}{2}\Big)}{\frac{h}{2}} \times \lim_{h \to 0} \cos\Big(\frac{2x-h}{2}\Big)$$
$$= 1 \times \cos x = \cos x $$

You can learn the proof much clearly here

http://www.mathdoubts.com/calculus/differentiation/identity/derivative-of-sinx/