Chapter 5, Problem 112RE

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 112RE

The required values of $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are 0.64, 0.70 and 1.18 respectively.

Explanation of Solution

Given information:

The identity: $\tan u=-\frac{\sqrt{45}}{2}$ , $\frac{\pi }{2}<u<\pi$

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,

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

Also, $\cos u=\frac{1}{\sec u}$

Here, $\sec u=6$

So, $\cos u=\frac{1}{6}=\frac{b}{h}$

This means, $b=1$ and $h=6$

And, $\sin u=\frac{p}{h}$

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

Substitute $1$ for $b$ and 6 for $h$ in the above formula and simplify for p.

  $\begin{array}{c}{\left(6\right)}^2=p^2+{\left(1\right)}^2\\ p^2=36-1\\ p^2=35\\ p=\pm \sqrt{35}\end{array}$

Since, sides cannot be negative; so, $p=\sqrt{35}$

Substitute $\sqrt{35}$ for $p$ and 6 for $h$ in the formula $\sin u=\frac{p}{h}$ .

  $\sin u=\frac{\sqrt{35}}{6}$

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

Substitute $\frac{1}{6}$ for $\cos u$ in the formula $\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\cos u}{2}}$ and simplify.

  $\begin{array}{c}\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\frac{1}{6}}{2}}\\ =\sqrt{\frac{\frac{6-1}{6}}{2}}\\ =\sqrt{\frac{5}{12}}\\ \approx 0.64\end{array}$

Substitute $\frac{1}{6}$ for $\cos u$ in the formula $\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\cos u}{2}}$ and simplify.

  $\begin{array}{c}\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\frac{1}{6}}{2}}\\ =\sqrt{\frac{\frac{6+1}{6}}{2}}\\ =\sqrt{\frac{7}{14}}\\ \approx 0.70\end{array}$

Substitute $\frac{1}{6}$ for $\cos u$ and $\frac{\sqrt{35}}{6}$ for $\sin u$ in the formula $\tan \left(\frac{u}{2}\right)=\frac{1-\cos u}{\sin u}$ and simplify.

  $\begin{array}{c}\tan \left(\frac{u}{2}\right)=\frac{1+\frac{1}{6}}{\frac{\sqrt{35}}{6}}\\ =\frac{\frac{6+1}{\cancel{6}}}{\frac{\sqrt{35}}{\cancel{6}}}\\ =\frac{7}{\sqrt{35}}\\ \approx 1.18\end{array}$

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 0.64, 0.70 and 1.18 respectively.