Derivative of Sine Function/Proof 5

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Theorem

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


Proof

\(\ds \map \arcsin x\) \(=\) \(\ds \int_0^x \frac {\d x} {\sqrt {1 - x^2} }\) Arcsin as an Integral
\(\ds \leadsto \ \ \) \(\ds \dfrac {\map \d {\map \arcsin y} } {\d y}\) \(=\) \(\ds \dfrac {\map \d {\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt {1 - y^2} }\) 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 $\text {III}$
\(\ds \dfrac {\map \d {\sin \theta} } {\d \theta}\) \(=\) \(\ds \cos \theta\)

$\blacksquare$


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