Chapter 5.5, Problem 42E

a.

To determine

To find the quadrant in which $\raisebox{1ex}{$u$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ lies.

Answer to Problem 42E

  $\raisebox{1ex}{$u$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ lies in I quadrant.

Explanation of Solution

Given:

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

Calculation:

  $\frac{\pi }{2}<u<\pi$

Dividing by 2 to the above inequality, we get

  $\begin{array}{l}\Rightarrow \frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}<\frac{u}{2}<\frac{\pi }{2}\\ \Rightarrow \frac{\pi }{4}<\frac{u}{2}<\frac{\pi }{2}\end{array}$

Therefore $\raisebox{1ex}{$u$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ lies in I quadrant.

b.

To determine

To simplify the given expression.

Answer to Problem 42E

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

Explanation of Solution

Given:

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

Concept used:

Pythagorean identity:

  ${\sin }^{2}u+{\cos }^{2}u=1$

Half-angle formulas:

  $\sin \frac{u}{2}=\pm \sqrt{\frac{1-\cos u}{2}},\cos \frac{u}{2}=\pm \sqrt{\frac{1+\cos u}{2}},\tan \frac{u}{2}=\frac{1-\cos u}{\sin u}$

The signs of $\sin \frac{u}{2}\&\cos \frac{u}{2}$ depend on the quadrant in which $\frac{u}{2}$ lies.

Quotient identity:

  $\tan u=\frac{\sin u}{\cos u}$

Calculation:

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

By using Pythagorean identity, we have

  $\begin{array}{l}{\cos }^{2}u=1-{\sin }^{2}u\\ \Rightarrow {\cos }^{2}u=1-{\left(\frac{5}{13}\right)}^2\\ \Rightarrow {\cos }^{2}u=1-\frac{25}{169}\\ \Rightarrow {\cos }^{2}u=\frac{144}{169}\\ \Rightarrow \cos u=\pm \frac{12}{13}\end{array}$

Since $\frac{\pi }{2}<u<\pi$ , $u$ lies in II quadrant. Moreover, $\cos u$ is negative in II quadrant.

  $\therefore \cos u=-\frac{12}{13}$

Now,

  $\begin{array}{l}\sin \frac{u}{2}=\pm \sqrt{\frac{1-\cos u}{2}}\\ \Rightarrow \sin \frac{u}{2}=\pm \sqrt{\frac{1-\left(-\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.\right)}{2}}=\pm \sqrt{\frac{25}{26}}=\pm \frac{5}{\sqrt{26}}\end{array}$

Since $\frac{u}{2}$ lies in quadrant I, moreover $\sin x$ is positive in quadrant I, therefore

  $\sin \frac{u}{2}=\frac{5}{\sqrt{26}}$

Now

  $\begin{array}{l}\cos \frac{u}{2}=\pm \sqrt{\frac{1+\cos u}{2}}\\ \Rightarrow \cos \frac{u}{2}=\pm \sqrt{\frac{1+\left(-\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.\right)}{2}}=\pm \sqrt{\frac{1}{26}}=\pm \frac{1}{\sqrt{26}}\end{array}$

Since $\frac{u}{2}$ lies in quadrant I, moreover $\cos x$ is positive in quadrant I, therefore

  $\cos \frac{u}{2}=\frac{1}{\sqrt{26}}$

Now by using quotient identity, we have

  $\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\sqrt{26}$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{26}$}\right.}=5$