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Derivative of the sine function

It can be shown that:

[math]\sin'(x)=\cos(x)[/math]

Prerequisite information[edit]

The proof[edit]

From the definition of the derivative:

[math]\begin{eqnarray} \sin'(x) &=& \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}\end{eqnarray}[/math]

Using the sine of sum formula [math]\sin(\alpha + \beta)=\sin \alpha \cos \beta +\sin \beta \cos \alpha[/math], we have

[math]\begin{eqnarray} \sin' (x) &=& \lim_{h\to 0} \frac{\sin (x)\cos (h) +\cos (x)\sin (h) -\sin (x)}{h}\end{eqnarray}[/math].

In the limit as $h\to 0$, we have $\sin h \to h$ and $\cos h \to 1$.

[math]\begin{eqnarray} \sin' (x) &=& \lim_{h\to 0} \frac{\sin (x) + h\cos (x) -\sin (x)}{h} \\ &=& \lim_{h\to 0} \frac{h\cos (x)}{h}\end{eqnarray}[/math]

Thus,

[math]\begin{eqnarray} \sin' (x) &=& \cos (x)\end{eqnarray}[/math]

Also see[edit]