Difference between revisions of "Derivative of sine"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
| − | $$\dfrac{d}{ | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ |
where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | From the definition, | |
| − | + | $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ | |
| + | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], and the definition of the cosine function, | ||
| + | $$\begin{array}{ll} | ||
| + | \dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) &= \dfrac{1}{2i} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] - \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ | ||
| + | &= \dfrac{1}{2i} \left[ ie^{iz} + ie^{-iz} \right] \\ | ||
| + | &= \dfrac{e^{iz}+e^{-iz}}{2} \\ | ||
| + | &= \cos(z), | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
| + | |||
| + | ==References== | ||
| + | *{{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=Derivative of cosine}}: $4.3.105$ | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 02:46, 5 January 2017
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
Proof
From the definition, $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the cosine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) &= \dfrac{1}{2i} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] - \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2i} \left[ ie^{iz} + ie^{-iz} \right] \\ &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \cos(z), \end{array}$$ as was to be shown. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.105$
