Difference between revisions of "Derivative of sine"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ | ||
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, | ||
$$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ | $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ | ||
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, | 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, | ||
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\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | + | ||
| − | + | ==References== | |
Revision as of 00:24, 4 June 2016
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. █
