Difference between revisions of "Derivative of cosine"

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==Theorem==
<strong>Proposition:</strong> $\dfrac{d}{dx}\cos(x) = -\sin(x)$
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The following formula holds:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) = -\sin(z),$$
<strong>Proof:</strong> █
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where $\cos$ denotes the [[cosine]] and $\sin$ denotes the [[sine]].
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==Proof==
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From the definition of cosine,
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$$\cos(z) = \dfrac{e^{iz}+e^{-iz}}{2},$$
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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]], the [[reciprocal of i]], and the definition of the sine function,
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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) &= \dfrac{1}{2} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] + \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\
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&= \dfrac{1}{2} \left[ ie^{iz} - ie^{-iz} \right] \\
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&= -\dfrac{e^{iz}-e^{-iz}}{2i} \\
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&= -\sin(z),
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\end{array}$$
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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=Derivative of sine|next=Derivative of tangent}}: $4.3.106$
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[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 01:27, 1 July 2017

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cos(z) = -\sin(z),$$ 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 reciprocal of 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