Derivative of Cosine Function/Proof 1
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Theorem
- $\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
Proof
From the definition of the cosine function, we have:
- $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
Then:
\(\ds \map {\frac \d {\d x} } {\cos x}\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^n 2 n \frac {x^{2 n - 1} } {\paren {2 n}!}\) | Power Series is Differentiable on Interval of Convergence | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\) | changing summation index | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}\) |
The result follows from the definition of the sine function.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (1) \ \text{(v)}$