Derivative of Sine Function/Proof 1
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Theorem
- $\map {\dfrac \d {\d x} } {\sin x} = \cos x$
Proof
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.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (1) \ \text{(vi)}$