How can I prove the derivative of $\sin(x)$ = $\cos(x)$ using two x values a and b - where b approaches a.

The first step here would be:

$$\lim_{a\to b}\frac{\sin (b) - \sin(a)}{b-a}$$

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    $\begingroup$ Are you allowed to use $\lim_{x\to 0}\frac{\sin x}{ x}=1$? $\endgroup$ – Andrei Oct 14 '19 at 20:00
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    $\begingroup$ Sum the sines. $\endgroup$ – conditionalMethod Oct 14 '19 at 20:02
  • $\begingroup$ @Andrei yes that's fine $\endgroup$ – hegash Oct 14 '19 at 20:11
  • $\begingroup$ @conditionalMethod Any more help on where to go from there? I found this problem in a textbook with no answers and have been working at it for about 3 hours. $\endgroup$ – hegash Oct 14 '19 at 20:12

I'm sorry but I'll use a notation that is more convenient for me to write in. Hope that's ok. The connection of course would be that $b=x$ and $a=x_0$

Consider the following: for every $x\neq x_o$ we have $$ \frac{\sin (x)-\sin \left(x_{0}\right)}{x-x_{0}}=\frac{2 \sin \left(\frac{x-x_{0}}{2}\right) \cdot \cos \left(\frac{x+x_{0}}{2}\right)}{x-x_{0}}=\frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}} \cdot \cos \left(\frac{x+x_{0}}{2}\right) $$

By the trig identity $ \sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) $. Now we substitute $y=\frac{x-x_{0}}{2}$, and see that: $$ \lim _{x \rightarrow x_{0}} \frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}}=\lim _{y \rightarrow 0} \frac{\sin y}{y}=1 $$ And: $$ \lim _{x \rightarrow x_{0}} \cos \left(\frac{x+x_{0}}{2}\right)=\cos x_{0} $$ After manipulation we arrived at limit of product, since both limits exist it'll be equal to the product of the limits, so: $$ \lim _{x \rightarrow x_{0}} \frac{\sin (x)-\sin \left(x_{0}\right)}{x-x_{0}}=\lim _{x \rightarrow x_{0}} \frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}} \cdot \cos \left(\frac{x+x_{0}}{2}\right)=1 \cdot \cos x_{0}=\cos{(x_0)} $$

As required.


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