Derivative of Sine Function/Proof 1

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Theorem

$\map {D_x} {\sin x} = \cos x$


Proof

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.

$\blacksquare$


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