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.
 A: $$f(x)=\sin x$$
Using first principle
$$f'(x)=\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$
$$=\lim_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)\cos\left(\frac{2x-h}{2}\right)}{\frac{h}{2}}$$
A: First question
$$\lim_{h\to0}\frac{\sin(x+h)-\sin x}h=\lim_{h\to0}\frac{\sin x\cos h+\sin h\cos x-\sin x}h=\cos x$$
You have to assume proved the limit $\lim_{h\to 0}\frac{\sin h}h=1$. The way to do this deppends completely on how you defined $\sin$.
Second question
A biconditional is true when both sides are true, and when both sides are false. In this case, both sides are true, so the biconditional is true.
A: Take
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\cos^2(x) + \sin^2(x)\right) = \frac{\mathrm{d}}{\mathrm{d}x} 1 = 0 \implies 2\cos(x)\cos(x)' + 2\sin(x)\sin(x)' = 0.  $$
Consequently,
$$ \frac{\cos(x)}{\sin(x)} = - \frac{\sin(x)'}{\cos(x)'} \quad \text{for } x \neq n \frac{\pi}{2}. $$
Hence, we must have that either
$$ \cos(x)' = -\sin(x) \quad \text{and} \quad \sin(x)' = \cos(x) $$
or
$$ \cos(x)' = \sin(x) \quad \text{and} \quad \sin(x)' = -\cos(x). $$
Geometrically, it is clear that as $x$ is increasing away from $0$ in the first quadrant, $\cos(x)$ is decreasing, i.e., $\cos(x)' < 0$. Hence, we must have that the first of the two alternatives above are correct, i.e.,
$$ \cos(x)' = - \sin(x) \quad \text{and} \quad \sin(x)' = \cos(x).  $$
The same argument can be repeated in each quadrant.
A: 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/
