Difference between revisions of "Derivative of sine"

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==Theorem==
<strong>[[Derivative of sine|Proposition]]:</strong> $\dfrac{d}{dx} \sin(x)=\cos(x)$
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The following formula holds:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$
<strong>Proof:</strong> █
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where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function.
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==Proof==
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From the definition,
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$$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$
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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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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) &= \dfrac{1}{2i} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] - \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\
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&= \dfrac{1}{2i} \left[ ie^{iz} + ie^{-iz} \right] \\
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&= \dfrac{e^{iz}+e^{-iz}}{2} \\
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&= \cos(z),
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\end{array}$$
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as was to be shown. █
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==References==
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*{{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=Derivative of cosine}}: $4.3.105$
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[[Category:Theorem]]
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[[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