Chapter 7.3, Problem 37E

To determine

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

Answer to Problem 37E

  $\sin \frac{x}{2}=\frac{1}{\sqrt{10}},\cos \frac{x}{2}=\frac{3}{\sqrt{10}}\&\tan \frac{x}{2}=\frac{1}{3}$

Explanation of Solution

Given:

  $\sin x=\frac{3}{5},0°<x<90°$

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:

Consider the given condition.

  $\sin x=\frac{3}{5},0°<x<90°$

By using Pythagorean identity, we have

  $\begin{array}{l}{\cos }^{2}x=1-{\sin }^{2}x\\ {\cos }^{2}x=1-{\left(\frac{3}{5}\right)}^2\\ {\cos }^{2}x=1-\frac{9}{25}\\ {\cos }^{2}x=\frac{16}{25}\\ \cos x=\pm \frac{4}{5}\end{array}$

Again, $0°<x<90°$

  $\cos x$ is positive

  $\cos x=\frac{4}{5}$

Consider the range of $x$

  $0°<x<90°$

Dividing by 2 to the above inequality, we get

  $\begin{array}{l}\frac{0°}{2}<\frac{x}{2}<\frac{90°}{2}\\ 0°<\frac{x}{2}<45°\end{array}$

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

  $\begin{array}{c}\sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}\\ =\pm \sqrt{\frac{1-\left(\frac{4}{5}\right)}{2}}\\ =\pm \sqrt{\frac{1}{10}}\\ =\pm \frac{1}{\sqrt{10}}\end{array}$

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

  $\sin \frac{x}{2}=\frac{1}{\sqrt{10}}$

Now

  $\begin{array}{c}\cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}\\ =\pm \sqrt{\frac{1+\left(\frac{4}{5}\right)}{2}}\\ =\pm \sqrt{\frac{9}{10}}\\ =\pm \frac{3}{\sqrt{10}}\end{array}$

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

  $\cos \frac{x}{2}=\frac{3}{\sqrt{10}}$

Now by using quotient identity,

  $\begin{array}{c}\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\\ =\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}\\ =\frac{1}{3}\end{array}$