Chapter 10.7, Problem 49E

Solution

To find: The formulae for $\tan \frac{\theta }{2}$ , $\sin \frac{\theta }{2}$ , and $\cos \frac{\theta }{2}$ .

The required formulae are:

  $\sin \left(\frac{\theta }{2}\right)=\sqrt{\left(\frac{1-\cos \theta }{2}\right)}$

  $\cos \left(\frac{\theta }{2}\right)=\sqrt{\frac{\left(1+\cos \theta \right)}{2}}$

  $tan\left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$

Given data: The diagram showing angles $\frac{\theta }{2}$ and $\theta$ .

  

Method/Formula used:

Every point on the circle is at distance equal to radius of the circle.

Calculation:

Let O be the center of the unit circle with radii OB and OA and ACB the right angle triangle.

Also, $OC=\cos \theta ,CB=\sin \theta$ , and $AC=1+\cos \theta$ .

In right angle triangle ABC ,

  $A{B}^2=A{C}^2+C{B}^2$

Substitute $1+\cos \theta$ for AC and $\sin \theta$ for CB,

  $\begin{array}{c}A{B}^2={\left(1+\cos \theta \right)}^2+{\left(\sin \theta \right)}^2\\ ={\cos }^2\theta +si{n}^2\theta +1+2\cos \theta \\ =1+1+2\cos \theta \leftarrow {\cos }^2\theta +si{n}^2\theta =1\\ =2\left(1+\cos \theta \right)\end{array}$

In triangle, ACB

  $\sin \left(\frac{\theta }{2}\right)=\frac{CB}{AB}$

Substitute $\sin \theta$ for CB and $\sqrt{2\left(1+\cos \theta \right)}$ for AB ,

  $\sin \left(\frac{\theta }{2}\right)=\frac{\sin \theta }{\sqrt{2\left(1+\cos \theta \right)}}$

Squaring both sides

  $\begin{array}{c}{\sin }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\theta }{2\left(1+\cos \theta \right)}\\ 2{\sin }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\theta }{\left(1+\cos \theta \right)}\frac{\left(1-\cos \theta \right)}{\left(1-\cos \theta \right)}\\ 2{\sin }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\theta \left(1-\cos \theta \right)}{\left(1-{\cos }^2\theta \right)}\leftarrow \left(1-{\cos }^2\theta \right)={\sin }^2\theta \\ 2{\sin }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\theta \left(1-\cos \theta \right)}{{\sin }^2\theta }\end{array}$

Solve further,

  $\left.\begin{array}{c}2{\sin }^2\left(\frac{\theta }{2}\right)=\frac{{\sin }^2\theta \left(1-\cos \theta \right)}{{\sin }^2\theta }\\ \sin \left(\frac{\theta }{2}\right)=\sqrt{\left(\frac{1-\cos \theta }{2}\right)}\end{array}\right\}$ … (1)

Also, in triangle, ACB

  $\cos \left(\frac{\theta }{2}\right)=\frac{CA}{AB}$

Substitute $1+\cos \theta$ for CB and $\sqrt{2\left(1+\cos \theta \right)}$ for AB ,

  $\cos \left(\frac{\theta }{2}\right)=\frac{1+\cos \theta }{\sqrt{2\left(1+\cos \theta \right)}}$

Squaring both sides

  $\left.\begin{array}{l}{\cos }^2\left(\frac{\theta }{2}\right)=\frac{{\left(1+\cos \theta \right)}^2}{2\left(1+\cos \theta \right)}\\ {\cos }^2\left(\frac{\theta }{2}\right)=\frac{\left(1+\cos \theta \right)}{2}\\ \cos \left(\frac{\theta }{2}\right)=\sqrt{\frac{\left(1+\cos \theta \right)}{2}}\end{array}\right\}$ … (2)

Further,

  $tan\left(\frac{\theta }{2}\right)=\frac{\sin \left(\frac{\theta }{2}\right)}{\cos \left(\frac{\theta }{2}\right)}$

Substitute $\sqrt{\left(\frac{1-\cos \theta }{2}\right)}$ for $\sin \left(\frac{\theta }{2}\right)$ and $\sqrt{\frac{\left(1+\cos \theta \right)}{2}}$ for $\cos \left(\frac{\theta }{2}\right)$

  $tan\left(\frac{\theta }{2}\right)=\frac{\sqrt{\left(\frac{1-\cos \theta }{2}\right)}}{\sqrt{\left(\frac{1+\cos \theta }{2}\right)}}$

Therefore,

  $tan\left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$