Chapter 5.5, Problem 38E

(a)

To determine

The Quadrant in which $\frac{u}{2}$ lies

Answer to Problem 38E

First Quadrant

Explanation of Solution

Given: $\sin u=\frac{5}{13},\frac{\pi }{2}<u<\pi$

The given expression is $\sin u=\frac{5}{13},\frac{\pi }{2}<u<\pi$

Divide by 2

  $\frac{\pi }{4}<\frac{u}{2}<\frac{\pi }{2}$

Here u is in 2^nd quadrant and

  $\frac{u}{2}$ lies in the First Quadrant

Hence, $\frac{u}{2}$ lies in First Quadrant

(b)

To determine

To Find: The exact values of $\sin \frac{u}{2},cos\frac{u}{2},\tan \frac{u}{2}$

Answer to Problem 38E

  $\sin \frac{u}{2}=\frac{5}{\sqrt{26}},\cos \frac{u}{2}=\frac{1}{\sqrt{26}},\tan \frac{u}{2}=5$

Explanation of Solution

Given:To find the exact values of $\sin \frac{u}{2},cos\frac{u}{2},\tan \frac{u}{2}$ using half angle formula

  $\sin u=\frac{5}{13},\frac{\pi }{2}<u<\pi$

FORMULA USED:

HALF ANGLE FORMULA:

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

The given expression is $\sin u=\frac{5}{13},\frac{\pi }{2}<u<\pi$

  $\begin{array}{l}base=-\sqrt{{13}^2-5^2}=-12\\ \cos u=\frac{base}{hyp}=\frac{-12}{13}\\ \sin \frac{u}{2}=\sqrt{\frac{1-\frac{(-12)}{13}}{2}}=\frac{5}{\sqrt{26}}\\ \cos \frac{u}{2}=\sqrt{\frac{1+\frac{(-12)}{13}}{2}}=\frac{1}{\sqrt{26}}\\ \tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\frac{5}{\sqrt{26}}.\frac{\sqrt{26}}{1}=5\end{array}$

All Sin , Cos, Tan are positive as the u/2 lies in First quadrant

Hence, $\sin \frac{u}{2}=\frac{5}{\sqrt{26}},\cos \frac{u}{2}=\frac{1}{\sqrt{26}},\tan \frac{u}{2}=5$