Chapter 5, Problem 109RE

To determine

Tocalculate:The exact values of $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ .

Answer to Problem 109RE

The required values of $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are 0.554, 0.832 and 0.667 respectively.

Explanation of Solution

Given information:

The identity: $\sin u=\frac{12}{13}$ , $0<u<\frac{\pi }{2}$

Formula used:

The half-angle formulas for trigonometric identities are shown below:

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

Pythagoras theorem:

For a right-angled triangle, the square of the hypotenuse and the sum of the squares of the base and the perpendicular are equal to each other that is,

  $\begin{array}{c}{\left(\text{hypotenuse}\right)}^2={\left(\text{perpendicular}\right)}^2+{\left(\text{base}\right)}^2\\ h^2=p^2+b^2\end{array}$

Calculation:

It is known that,

  $\sin \theta =\frac{\text{perpenidular}}{\text{hypotenuse}}=\frac{p}{h}$ and $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}=\frac{b}{h}$

Here, $\sin u=\frac{12}{13}=\frac{p}{h}$

This means, $p=12$ and $h=13$

And, $\cos u=\frac{b}{h}$

To find the base that is, b; use the Pythagoras theorem $h^2=p^2+b^2$ .

Substitute 12 for $p$ and 13 for $h$ in the above formula and simplify for $b$ .

  $\begin{array}{c}{\left(13\right)}^2={\left(12\right)}^2+b^2\\ b^2=169-144\\ b^2=25\\ b=\pm 5\end{array}$

Since, sides cannot be negative; so, $b=5$

Substitute 5 for $b$ and 13 for $h$ in the formula $\cos u=\frac{b}{h}$ .

  $\cos u=\frac{5}{13}$

To find the exact values, use the formulas $\sin \left(\frac{u}{2}\right)=\pm \sqrt{\frac{1-\cos u}{2}}$ ,

  $\cos \left(\frac{u}{2}\right)=\pm \sqrt{\frac{1+\cos u}{2}}$ and $\tan \left(\frac{u}{2}\right)=\frac{1-\cos u}{\sin u}$ .

Since, $0<u<\frac{\pi }{2}$ ; so $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are positive as they lie on the first quadrant.

Substitute $\frac{5}{13}$ for $\cos u$ in the formula $\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\cos u}{2}}$ and simplify.

  $\begin{array}{c}\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\frac{5}{13}}{2}}\\ =\sqrt{\frac{\frac{13-5}{13}}{2}}\\ =\sqrt{\frac{8}{26}}\\ \approx 0.554\end{array}$

Substitute $\frac{5}{13}$ for $\cos u$ in the formula $\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\cos u}{2}}$ and simplify.

  $\begin{array}{c}\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\frac{5}{13}}{2}}\\ =\sqrt{\frac{\frac{13+5}{13}}{2}}\\ =\sqrt{\frac{18}{26}}\\ \approx 0.832\end{array}$

Substitute $\frac{5}{13}$ for $\cos u$ and $\frac{12}{13}$ for $\sin u$ in the formula $\tan \left(\frac{u}{2}\right)=\frac{1-\cos u}{\sin u}$ and simplify.

  $\begin{array}{c}\tan \left(\frac{u}{2}\right)=\frac{1-\frac{5}{13}}{\frac{12}{13}}\\ =\frac{\frac{13-5}{\cancel{13}}}{\frac{12}{\cancel{13}}}\\ =\frac{8}{12}\\ \approx 0.667\end{array}$

Hence, the required values of $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are 0.554, 0.832 and 0.667 respectively.