Derivative of Sine Function

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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 $x$.


From Power Series is Differentiable on Interval of Convergence:

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


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
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\d \paren {\map \arcsin y} } {\d y}\) \(=\) \(\displaystyle \dfrac {\d \paren {\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\sqrt {1 - y^2} }\) 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 $\text {III}$
\(\displaystyle \dfrac {\d \paren {\sin \theta} } {\d \theta}\) \(=\) \(\displaystyle \cos \theta\)

$\blacksquare$


Also see


Sources