Chapter 13.9, Problem 32WE

To determine

To Find: The identity $\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }$ .

Answer to Problem 32WE

The right hand side is $\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }$ .

Explanation of Solution

The given identity is

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

Calculation:

Consider the identity is

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

Left hand side,

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

Using the half-angle formula for the sine and cosine, we get

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

Multiplying both the numerators and denominators by $(\sqrt{1-\cos \theta })$ ,

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

Applying the Pythagorean identity, we get

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

Therefore, the identity is proved $\tan \frac{\theta }{2}=\frac{1-\cos \theta }{\sin \theta }$ .