# Derivative of Sine Function

## Theorem

$\map {D_x} {\sin x} = \cos x$

### Corollary

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

## 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 Radius of Convergence of Power Series over Factorial, this series converges for all $x$.

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

The result follows from the definition of the cosine function.

$\blacksquare$

## Proof 2

 $\displaystyle D_x \left({\sin x}\right)$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({x + h}\right) - \sin \left({x}\right)} h$ Definition of Derivative of Real Function at Point $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({x}\right) \cos \left({h}\right) + \sin \left({h}\right) \cos \left({x}\right) - \sin \left({x}\right)} h$ Sine of Sum $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) + \sin \left({h}\right) \cos \left({x}\right)} h$ collecting terms containing $\sin \left({x}\right)$ and factoring $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({x}\right) \left({\cos \left({h}\right) - 1}\right) } h + \lim_{h \mathop \to 0} \frac {\sin \left({h}\right) \cos \left({x}\right)} h$ Sum Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle \sin \left({x}\right) \times 0 + 1 \times \cos \left({x}\right)$ Limit of Sine of X over X and Limit of (Cosine (X) - 1) over X $\displaystyle$ $=$ $\displaystyle \cos \left({x}\right)$

$\blacksquare$

## Proof 3

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

$\blacksquare$

## Proof 4

 $\displaystyle D_x \left({\sin x}\right)$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({x + h}\right) - \sin \left({x}\right)} h$ Definition of Derivative of Real Function at Point $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\sin \left({ \left({x + \frac h 2}\right) + \frac h 2}\right) - \sin \left({ \left({x + \tfrac h 2}\right) - \tfrac h 2}\right)} h$ $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {2 \cos \left({x + \frac h 2}\right) \sin \left({\frac h 2}\right)} h$ Simpson's Formula for Cosine by Sine $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \cos \left({x + \frac h 2}\right) \lim_{h \mathop \to 0} \frac{\sin \left({\frac h 2}\right)} {\frac h 2}$ Multiple Rule for Limits of Functions and Product Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle \cos x \times 1$ Continuity of Cosine and Limit of Sine of X over X $\displaystyle$ $=$ $\displaystyle \cos x$

$\blacksquare$

## Proof 5

 $\displaystyle \map \arcsin x$ $=$ $\displaystyle \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ Arcsin as an Integral $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\d \paren {\map \arcsin y} } {\d y}$ $=$ $\displaystyle \dfrac {\d \paren \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {\sqrt {1 - y^2} }$ Corollary to Fundamental Theorem of Calculus: First Part

Note that we get the same answer as Derivative of Arcsine Function.

By definition of real $\arcsin$ function, $\arcsin$ is bijective on its domain $\closedint {-1} 1$.

Thus its inverse is itself a mapping.

From Inverse of Inverse of Bijection, its inverse is the $\sin$ function.

 $\displaystyle \dfrac {\d \paren {\sin \theta} } {\d \theta}$ $=$ $\displaystyle 1 / \dfrac 1 {\sqrt {1 - \sin^2 \theta} }$ Derivative of Inverse Function $\displaystyle$ $=$ $\displaystyle \pm \sqrt {1 - \sin^2 \theta}$ Positive in Quadrant $\text I$ and Quadrant $\text {IV}$, Negative in Quadrant $\text {II}$ and Quadrant $\text {III}$ $\displaystyle \dfrac {\d \paren {\sin \theta} } {\d \theta}$ $=$ $\displaystyle \cos \theta$

$\blacksquare$