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Using $$ \lim_{x \to 0} \frac{\sin{x}}{x} = 1 $$ and summation theorem for sine, prove: $(\sin{x})^{'} = \cos{x}$

So I wrote: $$ \lim_{h \to 0} \frac{\sin{(x+h)} - \sin{x}}{h} = \lim_{h \to 0} \frac{2\sin{(\frac{h}{2})}\cos{(x + \frac{h}{2})}}{h} = \lim_{h \to 0} \frac{\sin{(\frac{h}{2})}}{\frac{h}{2}} \cdot \lim_{h \to 0} \cos{(x+ \frac{h}{2})} = \cos{x}$$ Is that right? Because I don't know what summation theorem for sine is.

Samuel M. A. Luque
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NightEye
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    Thsi is correct. Perhaps they want you to use the formula $\sin (x+h)=\sin x\cos h+\cos x \sin h$ but what you have done is better. – Kavi Rama Murthy Jun 10 '21 at 09:05
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    It's right , but according to standered defination of differentiation , we takes two point in the domain of function . $t$ and $x$. Or t= $x+h$. And while forming quotient , x+h approaches to x. So overall i want to say that that h will approach to 0 in limit ( not x ) –  Jun 10 '21 at 09:10
  • thanks for comments – NightEye Jun 10 '21 at 09:15
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    You should use \sin and \cos for better formatting of the functions. Also, use \left( and \right) instead of ( and ) in the last line to adjust that bracket size. – ultralegend5385 Jun 10 '21 at 09:16
  • Mh, how can you claim that you don't know a theorem and use it at the same time ? –  Jun 10 '21 at 09:29
  • @KaviRamaMurthy: How do you show that $$\sin(x+h)-\sin(x)=2\sin\left(\frac{h}{2}\right)\cos\left(\frac{h}{2}+x\right) , ?$$Which identities is this using? – Joe Jun 10 '21 at 10:11
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    @KaviRamaMurthy $\sin (a) - sin(b) = 2 \sin \frac{1}{2} (a - b) \cdotp \cos \frac{1}{2}(a+b) $ – NightEye Jun 10 '21 at 11:00

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Your proof is better, and is infact better than the sum identity.

Firstly, you put $x\to 0$ in the limit where it should be $h\to 0$. Probably a typo, but wanted to correct that.

Second, you should write $\frac h2\to 0$ where you use the limit identity. This is not something needed in later studies, but at this level (presuming on the level of the problem), you should write it as a good practice.

Hope this helps. Ask anything if not clear :)

ultralegend5385
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