# Derivative of Sine Function

## Theorem

$D_x \left({\sin x}\right) = \cos x$

### Corollary

$D_x \left({\sin \left({a x}\right)}\right) = a \cos \left({a x}\right)$

## Proof 1

From the definition of the sine function, we have:

$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1}} {\left({2n+1}\right)!}$

From Power Series over Factorial, this series converges for all $x$.

From Power Series Differentiable on Interval of Convergence, we have:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle D_x \left({\sin x}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \left({2n+1}\right) \frac {x^{2n} } {\left({2n+1}\right)!}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n} } {\left({2n}\right)!}$$ $$\displaystyle$$ $$\displaystyle$$

The result follows from the definition of the cosine function.

$\blacksquare$

## Proof 2

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle D_x \left({\sin x}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lim_{h \to 0} \frac { \sin \left({x + h}\right) - \sin \left({x}\right) } h$$ $$\displaystyle$$ $$\displaystyle$$ Definition of derivative at a point $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \cos \left({h}\right) + \sin \left({h}\right) \cos \left({x}\right) - \sin \left({x}\right) } h$$ $$\displaystyle$$ $$\displaystyle$$ Sine of Sum $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) + \sin \left({h}\right) \cos \left({x}\right) } h$$ $$\displaystyle$$ $$\displaystyle$$ Collecting terms containing $\sin \left({x}\right)$ and factoring $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lim_{h \to 0} \frac { \sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) } h + \lim_{h \to 0} \frac { \sin \left({h}\right) \cos \left({x}\right) } h$$ $$\displaystyle$$ $$\displaystyle$$ Sum Rule $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sin \left({x}\right) \times 0 + 1 \times \cos \left({x}\right)$$ $$\displaystyle$$ $$\displaystyle$$ From Limit of Sine of X over X and Limit of (Cosine (X) - 1) over X $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \cos \left({x}\right)$$ $$\displaystyle$$ $$\displaystyle$$

$\blacksquare$

## Proof 3

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle D_x \sin x$$ $$=$$ $$\displaystyle$$ $$\displaystyle D_x \cos \left({\frac \pi 2 - x}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Cosine of Complement equals Sine $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \sin \left({\frac \pi 2 - x}\right)$$ $$\displaystyle$$ $$\displaystyle$$ Derivative of Cosine Function and Chain Rule $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \cos x$$ $$\displaystyle$$ $$\displaystyle$$ Sine of Complement equals Cosine

$\blacksquare$