Derivative of Sine Function/Proof 5
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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 $\text {III}$ | |||||||||||
\(\ds \dfrac {\map \d {\sin \theta} } {\d \theta}\) | \(=\) | \(\ds \cos \theta\) |
$\blacksquare$
Sources
- robjohn (https://math.stackexchange.com/users/13854/robjohn), How to prove that $\ds \lim \limits_{x \mathop \to 0}\frac {\sin x} x = 1$?, URL (version: 2013-06-19): https://math.stackexchange.com/q/75151