Difference between revisions of "Derivative of cosine"
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==Proof== | ==Proof== | ||
| + | From the definition of cosine, | ||
| + | $$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$ | ||
| + | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], the fact that $\dfrac{1}{i}=-i$, 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== | ==References== | ||
Revision as of 03:37, 30 June 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cos(x) = -\sin(x),$$ 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 fact that $\dfrac{1}{i}=-i$, 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. █
