Derivative of Sine Function/Proof 1

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


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 $x$.

From Power Series is Differentiable on Interval of Convergence:

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

The result follows from the definition of the cosine function.