# Derivative of sin(x) = cos(x) from first principles without using (a + h)

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:

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

• Are you allowed to use $\lim_{x\to 0}\frac{\sin x}{ x}=1$? – Andrei Oct 14 '19 at 20:00
• – conditionalMethod Oct 14 '19 at 20:02
• @Andrei yes that's fine – hegash Oct 14 '19 at 20:11
• @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. – 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: $$\lim _{x \rightarrow x_{0}} \frac{\sin (x)-\sin \left(x_{0}\right)}{x-x_{0}}=\lim _{x \rightarrow x_{0}} \frac{\sin \left(\frac{x-x_{0}}{2}\right)}{\frac{x-x_{0}}{2}} \cdot \cos \left(\frac{x+x_{0}}{2}\right)=1 \cdot \cos x_{0}=\cos{(x_0)}$$