Last modified on 27 November 2014, at 01:47

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\left(\theta+\frac{h}{2}\right)}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}.

Clearly, the limit of the first term is \cos(\theta) since \cos(\theta) is a continuous function. Write k = h2; the second term is then


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).