Chapter 13.4, Problem 41HP

To determine

To derive: the half −angle identities from the double −angle identities.

Answer to Problem 41HP

  $\begin{array}{l}\sin \theta =\pm \sqrt{\frac{1-\cos 2\theta }{2}}.\\ \cos \theta =\pm \sqrt{\frac{1+\cos 2\theta }{2}}.\end{array}$

Explanation of Solution

Given information: Have to derive the half- angle identities from the double −angle

identities.

Formula Used:

  $\cos 2\theta =1-2{\sin }^2\theta$ .

  $\cos 2\theta =2{\cos }^2\theta -1$ .

Calculation:

To find the half - angle identity for $\sin \theta$ , start from the double − angle identity:

  $\begin{array}{c}\cos 2\theta =1-2{\sin }^2\theta \\ \cos 2\theta -1=-2{\sin }^2\theta \\ 2{\sin }^2\theta =1-\cos 2\theta \\ {\sin }^2\theta =\frac{1-\cos 2\theta }{2}\\ \sin \theta =\pm \sqrt{\frac{1-\cos 2\theta }{2}}\end{array}$

To find the half - angle identity for $\cos \theta$ , start from the double − angle identity:

  $\begin{array}{c}\cos 2\theta =2{\cos }^2\theta -1\\ \cos 2\theta +1=2{\cos }^2\theta \\ 2{\cos }^2\theta =1+\cos 2\theta \\ {\cos }^2\theta =\frac{1+\cos 2\theta }{2}\\ \cos \theta =\pm \sqrt{\frac{1+\cos 2\theta }{2}}\end{array}$

Hence derived.

  $\begin{array}{l}\sin \theta =\pm \sqrt{\frac{1-\cos 2\theta }{2}}.\\ \cos \theta =\pm \sqrt{\frac{1+\cos 2\theta }{2}}.\end{array}$