Trigonometry/Derivative of Cosine

      To find the derivative of cos(θ).

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

      As in the proof of Derivative of Sine, the limit of the first term is \displaystyle\sin(\theta) and the limit of the second term is 1. Thus

      \frac{d}{d\theta} \cos(\theta) = -\sin(\theta).


      Thus

      \frac{d}{d\theta} \sin(\theta) = \cos(\theta)
      \frac{d}{d\theta} \cos(\theta) = -\sin(\theta)
      \frac{d}{d\theta} -\sin(\theta) = -\cos(\theta)
      \frac{d}{d\theta} -\cos(\theta) = \sin(\theta)

      and so on for ever.

      Last modified on 15 March 2011, at 17:44