# Derivative of Sine Function

## Theorem

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

### Corollary

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

## Proof 1

From the definition of the sine function, we have:

$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$

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

 $\ds \map {\frac \d {\d x} } {\sin x}$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$

The result follows from the definition of the cosine function.

$\blacksquare$

## Proof 2

 $\ds \map {\frac \d {\d x} } {\sin x}$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\sin x \cos h + \sin h \cos x - \sin x} h$ Sine of Sum $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} + \sin h \cos x} h$ collecting terms containing $\map \sin x$ and factoring $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} } h + \lim_{h \mathop \to 0} \frac {\sin h \cos x} h$ Sum Rule for Limits of Real Functions $\ds$ $=$ $\ds \map \sin x \times 0 + 1 \times \cos x$ Limit of $\dfrac {\sin x} x$ at Zero and Limit of $\dfrac {\cos x - 1} x$ at Zero $\ds$ $=$ $\ds \cos x$

$\blacksquare$

## Proof 3

 $\ds \dfrac \d {\d x} \sin x$ $=$ $\ds \dfrac \d {\d x} \map \cos {\frac \pi 2 - x}$ Cosine of Complement equals Sine $\ds$ $=$ $\ds \map \sin {\frac \pi 2 - x}$ Derivative of Cosine Function and Chain Rule for Derivatives $\ds$ $=$ $\ds \cos x$ Sine of Complement equals Cosine

$\blacksquare$

## Proof 4

 $\ds \map {\frac \d {\d x} } {\sin x}$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - \frac h 2} } h$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {2 \map \cos {x + \frac h 2} \map \sin {\frac h 2} } h$ Simpson's Formula for Cosine by Sine $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \map \cos {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}$ Multiple Rule for Limits of Real Functions and Product Rule for Limits of Real Functions $\ds$ $=$ $\ds \cos x \times 1$ Cosine Function is Continuous and Limit of $\dfrac {\sin x} x$ at Zero $\ds$ $=$ $\ds \cos x$

$\blacksquare$

## Proof 5

 $\ds \map \arcsin x$ $=$ $\ds \int_0^x \frac {\d x} {\sqrt {1 - x^2} }$ 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 $\text {III}$ $\ds \dfrac {\map \d {\sin \theta} } {\d \theta}$ $=$ $\ds \cos \theta$

$\blacksquare$