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),$$ |
where $\cos$ denotes the [[cosine]] and $\sin$ denotes the [[sine]]. | where $\cos$ denotes the [[cosine]] and $\sin$ denotes the [[sine]]. | ||
| Line 7: | Line 7: | ||
From the definition of cosine, | From the definition of cosine, | ||
$$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ | $$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ | ||
| − | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], the | + | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], the [[reciprocal of i]], and the definition of the sine function, |
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ | \dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ | ||
| Line 14: | Line 14: | ||
&= -\sin(z), | &= -\sin(z), | ||
\end{array}$$ | \end{array}$$ | ||
| − | as was to be shown. █ | + | as was to be shown. █ |
==References== | ==References== | ||
| + | *{{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Derivative of sine|next=Derivative of tangent}}: $4.3.106$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
Latest revision as of 01:27, 1 July 2017
Theorem
The following formula holds: $$\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): $4.3.106$
