Derivative of Sine Function

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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:

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

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


From Power Series is Differentiable on Interval of Convergence:

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


The result follows from the definition of the cosine function.

$\blacksquare$


Proof 2

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

$\blacksquare$


Proof 3

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

$\blacksquare$


Proof 4

\(\displaystyle \map {\frac \d {d x} } {\sin x}\) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h\) Definition of Derivative of Real Function at Point
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - \frac h 2} } h\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {2 \map \cos {x + \frac h 2} \map \sin {\frac h 2} } h\) Simpson's Formula for Cosine by Sine
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \map \cos {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}\) Multiple Rule for Limits of Real Functions and Product Rule for Limits of Real Functions
\(\displaystyle \) \(=\) \(\displaystyle \cos x \times 1\) Cosine Function is Continuous 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
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\map \d {\map \arcsin y} } {\d y}\) \(=\) \(\displaystyle \dfrac {\map \d {\displaystyle \int_0^y \dfrac 1 {\sqrt {1 - y^2} } \rd y} } {\d y}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\sqrt {1 - y^2} }\) 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.

\(\displaystyle \dfrac {\map \d {\sin \theta} } {\d \theta}\) \(=\) \(\displaystyle 1 / \dfrac 1 {\sqrt {1 - \sin^2 \theta} }\) 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 $\text {III}$
\(\displaystyle \dfrac {\map \d {\sin \theta} } {\d \theta}\) \(=\) \(\displaystyle \cos \theta\)

$\blacksquare$


Also see


Sources