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:  
$$\dfrac{d}{dx} \sin(x) = \cos(x),$$
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$
 
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> █  
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<strong>Proof:</strong> 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 [[linear property of derivative 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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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. █