Chapter 4.1, Problem 57E

To determine

To prove: $\frac{d}{dx}\cos x^{\circ }\text{ is equal to}-\frac{\pi }{180}\sin x^{\circ }$ .

Answer to Problem 57E

  $\frac{d}{dx}\cos x^{\circ }\text{ is equal to}-\frac{\pi }{180}\sin x^{\circ }$

Explanation of Solution

Given information:

  $\frac{d}{dx}\cos x^{\circ }=-\frac{\pi }{180}\sin x^{\circ }$

To convert $x^{\circ }$ to radians multiply $\frac{\pi }{180}$ ,

  $\begin{array}{c}\frac{d}{dx}\cos x^{\circ }=\frac{d}{dx}\cos \frac{\pi x^{\circ }}{180}\\ \frac{d}{dx}\cos \frac{\pi x^{\circ }}{180}=-\sin \frac{\pi x^{\circ }}{180}\times \frac{\pi }{180}\to \text{by using chain rule}\\ \frac{d}{dx}\cos x^{\circ }=-\frac{\pi }{180}\sin x^{\circ }\end{array}$

Hence it has been proved that $\frac{d}{dx}\cos x^{\circ }\text{ is equal to}-\frac{\pi }{180}\sin x^{\circ }$ .