Trigonometry/Derivative of Sine

To find the derivative of sin(θ).

\frac{d}{d\theta} \sin(\theta) = \lim_{h \rightarrow 0} \frac{\sin(\theta+h)-\sin(\theta)}{h} = \lim_{h \rightarrow 0} \frac{2\cos(\theta+\frac{h}{2})\sin(\frac{h}{2})}{h} = \lim_{h \rightarrow 0} {\cos(\theta+\frac{h}{2})}\frac{\sin(\frac{h}{2})}{\frac{h}{2}}.

Clearly, the limit of the first term is \cos(\theta) since \cos(\theta) is a continuous function. Write k = h⁄2; the second term is then

\frac{\sin(k)}{k}.

Which we proved earlier tends to 1 as k \rightarrow 0.

And since

k \rightarrow 0 \text { as } h \rightarrow 0,

the limit of the second term is 1 too. Thus

\frac{d}{d\theta} \sin(\theta) = \cos(\theta).
Last modified on 1 April 2011, at 15:40