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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.

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  • $\begingroup$ Use mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $$\sin(x+h)-\sin x$$ $\endgroup$ – lab bhattacharjee Aug 16 '16 at 16:50
  • $\begingroup$ Thank you. What about secand question? $\endgroup$ – Jamal Farokhi Aug 16 '16 at 16:52
  • $\begingroup$ Of course that proposition is true. Both sides of the implication are true! What you seem to want is to use the latter in a proof. This is a good idea. $\endgroup$ – Jacob Wakem Aug 17 '16 at 1:55
  • $\begingroup$ You mean sin^2 (x) not sin (x)^2. Similarly for cos. $\endgroup$ – Jacob Wakem Aug 17 '16 at 1:57
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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.

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  • $\begingroup$ You also need $\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$. $\endgroup$ – user361424 Aug 18 '16 at 3:42
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$$f(x)=\sin x$$ Using first principle $$f'(x)=\lim_{h\to 0}\frac{\sin(x+h) -\sin x}{h}$$ $$=\lim_{h\to 0}\frac{\sin\left(\frac{h}{2}\right)\cos\left(\frac{2x-h}{2}\right)}{\frac{h}{2}}$$

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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/

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