# Derivative of Sine Function/Proof 5

## Theorem

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$

## Proof

 $\ds \map \arcsin x$ $=$ $\ds \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ Arcsine as Integral $\ds \leadsto \ \$ $\ds \dfrac {\map \d {\map \arcsin y} } {\d y}$ $=$ $\ds \dfrac {\map \d {\ds \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt {1 - y^2} }$ 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.

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

$\blacksquare$