Difference between revisions of "Derivative of sine"
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<strong>[[Derivative of sine|Proposition]]:</strong> The following formula holds: | <strong>[[Derivative of sine|Proposition]]:</strong> The following formula holds: | ||
− | $$\dfrac{d}{ | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ |
where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | ||
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− | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> From the definition, |
+ | $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ | ||
+ | and so using the [[derivative of the exponential function]], the [[linear property of derivative operator|linear property of the derivative]], the [[chain rule]], and the definition of the cosine function, | ||
+ | $$\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}$$ | ||
+ | as was to be shown. █ | ||
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Revision as of 05:34, 8 February 2016
Proposition: The following formula holds: $$\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. █