I have to prove that $\dfrac{d}{dx}\sin(x)=\cos(x)$. I used the definition of a derivative: $$\dfrac{d}{dx}f(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$ $$\dfrac{d}{dx}\sin(x)=\lim\limits_{h\to 0} \dfrac{\sin(x+h)-\sin(x)}{h}$$ Using angle sum formula: $$\dfrac{d}{dx}\sin(x)=\lim\limits_{h\to 0} \dfrac{\sin (x)\cos (h)+\cos (x)\sin (h)-\sin (x)}{h}$$ Rearranging the terms: $$\dfrac{d}{dx}\sin(x)=\lim\limits_{h\to 0} \dfrac{\cos (h)\sin(h)-\sin(x)+\sin(x)\cos(h)}{h}$$ Factoring out $\sin(x)$ from the last two terms: $$\dfrac{d}{dx}\sin(x)=\lim\limits_{h\to 0} \dfrac{\cos(x)\sin(h)-\sin(x)[1-\cos(h)]}{h}$$ Separating the fraction: $$\dfrac{d}{dx}\sin(x)=\lim\limits_{h\to 0} \dfrac{\cos(x)\sin(h)}{h}-\lim\limits_{h\to 0} \dfrac{\sin(x)[1-\cos(h)]}{h}$$ Now I am stuck. Can anyone please tell me what to do next, or give me a hint? Thanks
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$\begingroup$Now it depends on what you know. Do you have $\lim_{h \to 0} \frac {\sin h}h=1$? And $\lim_{h \to 0} \frac {1-\cos h}h=0$? Then insert them and you are done. These usually come before what you are doing. The Taylor series are another route, but usually come later.
$\endgroup$ 2 $\begingroup$See this answer for a geometric proof that $$ \lim_{x\to0}\frac{\sin(x)}{x}=1 $$ You might also find the following useful: $$ \frac{1-\cos(x)}{x}=\frac{\sin(x)}{x}\frac{\sin(x)}{1+\cos(x)} $$
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