Chapter 7.4, Problem 16E

To determine

To find: the exact value of the function by using the half-angle identity.

Answer to Problem 16E

  $2+\sqrt{3}$

Explanation of Solution

Given information:

  $\tan \frac{5\pi }{12}$

Calculation:

Evaluate $\alpha$ as shown below.

  $\begin{array}{c}\frac{\alpha }{2}=\frac{5\pi }{12}\\ \alpha =2\left(\frac{5\pi }{12}\right)\\ =\frac{5\pi }{6}\end{array}$

So the value of $\alpha$ is $\frac{5\pi }{6}$ .

Solve the function by using half-angle identity for tangent as shown below.

  $\begin{array}{c}\tan \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}\\ \tan \frac{5\pi }{12}=\sqrt{\frac{1-\cos \left(\raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}{1+\cos \left(\raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)}}\left[\begin{array}{l}\text{Put }\alpha =\raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\\ \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$12$}\right.\text{ lies in Quadrant I, so }\tan \raisebox{1ex}{$5\pi $}\!\left/ \!\raisebox{-1ex}{$12$}\right.\text{ is positive}\end{array}\right]\\ =\sqrt{\frac{1-\left(\raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{1+\left(\raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}\\ =\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\\ =\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}\left(\frac{2+\sqrt{3}}{2+\sqrt{3}}\right)}\left[\text{Multiply and divide by }2+\sqrt{3}\right]\\ =\sqrt{\frac{{\left(2+\sqrt{3}\right)}^2}{2^2-{\left(\sqrt{3}\right)}^2}}\left[a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\\ =\sqrt{\frac{{\left(2+\sqrt{3}\right)}^2}{4-3}}\\ =2+\sqrt{3}\end{array}$

Thus the value of given function is $2+\sqrt{3}$ .