Difference between revisions of "Derivative of cosine"
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$\dfrac{\mathrm{d}}{\mathrm{d} | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) = -\sin(z),$$ |
where $\cos$ denotes the [[cosine]] and $\sin$ denotes the [[sine]]. | where $\cos$ denotes the [[cosine]] and $\sin$ denotes the [[sine]]. | ||
Latest revision as of 01:27, 1 July 2017
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) = -\sin(z),$$ where $\cos$ denotes the cosine and $\sin$ denotes the sine.
Proof
From the definition of cosine, $$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, the reciprocal of i, and the definition of the sine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2} \left[ ie^{iz} - ie^{-iz} \right] \\ &= -\dfrac{e^{iz}-e^{-iz}}{2i} \\ &= -\sin(z), \end{array}$$ as was to be shown. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.106$
