Chapter 5, Problem 8PS

To determine

The formula for $\sin \frac{\theta }{2},\cos \frac{\theta }{2},\text{and}\tan \frac{\theta }{2}$

Answer to Problem 8PS

  $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ ; $\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

  $\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }$

Explanation of Solution

Given information:

The given derived formula as following below,

  $\sin \frac{\theta }{2},\cos \frac{\theta }{2},\text{and}\tan \frac{\theta }{2}$

Formula used:

  $\tan \frac{\theta }{2}=\frac{\text{Opposite Side}}{\text{Adjacent Side}}$ ; $\cos \frac{\theta }{2}=\frac{\text{Adjacent Side}}{\text{Hypotenuse}}$ ; $\sin \frac{\theta }{2}=\frac{\text{opposite Side}}{\text{Hypotenuse}}$

Calculation:

Use following trigonometric identity:

  $\sin \theta =2\sin \frac{\theta }{2}\times \cos \frac{\theta }{2}$

This can be written as

  $\sin \frac{\theta }{2}=\frac{\sin \theta }{2\cos \frac{\theta }{2}}$

Further use identity:

  $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$

Therefore, value is given by

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

The sign of $\sin \frac{\theta }{2}$ depends upon the quadrant in which $\frac{\theta }{2}$ lies.

Similarly, value of $\cos \frac{\theta }{2}$ is given by

  $\begin{array}{l}\cos \frac{\theta }{2}=\frac{\text{Adjacent Side}}{\text{Hypotenuse}}\\ =\frac{1+\cos \theta }{\sqrt{{\left(1+\cos \theta \right)}^2+{\left(\sin \theta \right)}^2}}\\ =\frac{1+\cos \theta }{\sqrt{1+{\cos }^2\theta +2\cos \theta +{\sin }^2\theta }}\\ \cos \frac{\theta }{2}=\frac{1+\cos \theta }{\sqrt{2+2\cos \theta }}\end{array}$

On simplification,

  $\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

The sign of $\cos \frac{\theta }{2}$ depends upon the quadrant in which $\frac{\theta }{2}$ lies

Similarly, value of $\tan \frac{\theta }{2}$ is given by

  $\begin{array}{l}\tan \frac{\theta }{2}=\frac{\text{Opposite Side}}{\text{Adjacent Side}}\\ \tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }\end{array}$

Conclusion:

The values are$\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ ; $\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

  $\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }$