Chapter 7.3, Problem 42E

To determine

To find the values of $\sin \frac{x}{2},\cos \frac{x}{2}\&\tan \frac{x}{2}$ .

Answer to Problem 42E

  $\sin \frac{x}{2}=\sqrt{\frac{\sqrt{26}+5}{2\sqrt{26}}},\cos \frac{x}{2}=-\sqrt{\frac{\sqrt{26}-5}{2\sqrt{26}}}\&\tan \frac{x}{2}=-\left(\sqrt{26}+5\right)$

Explanation of Solution

Given:

  $\cot x=5,180°<x<270°$

Concept used:

Pythagorean identity:

  $1+{\tan }^{2}u={\sec }^{2}u$

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:

  $\tan u=\frac{\sin u}{\cos u}$

Reciprocal identity:

  $\begin{array}{l}\sec u=\frac{1}{\cos u}\\ \cot u=\frac{1}{\tan u}\end{array}$

Calculation:

Consider the given condition.

  $\cot x=5,180°<x<270°$

So,

  $\tan x=\frac{1}{5}$

By using Pythagorean identity, we have

  $\begin{array}{l}{\sec }^{2}x=1+{\tan }^{2}x\\ {\sec }^{2}x=1+{\left(\frac{1}{5}\right)}^2\\ {\sec }^{2}x=1+\frac{1}{25}\\ {\sec }^{2}x=\frac{26}{25}\\ \sec x=\pm \frac{\sqrt{26}}{5}\\ \cos x=\pm \frac{5}{\sqrt{26}}\end{array}$

Again, $180°<x<270°$

  $\cos x$ is negative

  $\cos x=-\frac{5}{\sqrt{26}}$

Consider the range of $x$

  $180°<x<270°$

  $\begin{array}{l}\frac{180°}{2}<\frac{x}{2}<\frac{270°}{2}\\ 90°<\frac{x}{2}<135°\end{array}$

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

  $\begin{array}{c}\sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}\\ =\pm \sqrt{\frac{1-\left(-\frac{5}{\sqrt{26}}\right)}{2}}\\ =\pm \sqrt{\frac{\sqrt{26}+5}{2\sqrt{26}}}\end{array}$

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

  $\sin \frac{x}{2}=\sqrt{\frac{\sqrt{26}+5}{2\sqrt{26}}}$

Now

  $\begin{array}{c}\cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}\\ =\pm \sqrt{\frac{1+\left(-\frac{5}{\sqrt{26}}\right)}{2}}\\ =\pm \sqrt{\frac{\sqrt{26}-5}{2\sqrt{26}}}\end{array}$

Since $\frac{x}{2}$ lies in quadrant II, moreover $\cos x$ is negative in quadrant II, therefore

  $\cos \frac{x}{2}=-\sqrt{\frac{\sqrt{26}-5}{2\sqrt{26}}}$

Now by using quotient identity,

  $\begin{array}{c}\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\\ =-\frac{\sqrt{\frac{\sqrt{26}+5}{2\sqrt{26}}}}{\sqrt{\frac{\sqrt{26}-5}{2\sqrt{26}}}}\\ =-\sqrt{\frac{\sqrt{26}+5}{\sqrt{26}-5}}\times \sqrt{\frac{\sqrt{26}-5}{\sqrt{26}-5}}\\ =-\frac{1}{\sqrt{26}-5}\\ =-\frac{1}{\sqrt{26}-5}\times \frac{\sqrt{26}+5}{\sqrt{26}+5}\\ =-\frac{\sqrt{26}+5}{26-25}\\ =-\left(\sqrt{26}+5\right)\end{array}$