Chapter 5, Problem 111RE

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 111RE

The required values of $\sin \left(\frac{u}{2}\right)$ , $\cos \left(\frac{u}{2}\right)$ and $\tan \left(\frac{u}{2}\right)$ are 0.59, 0.8 and 0.74 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 $\tan u=\frac{p}{b}$

Here, $\tan u=-\frac{\sqrt{45}}{2}=\frac{p}{b}$

This means, $p=-\sqrt{45}$ and $b=2$

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

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

Substitute $-\sqrt{45}$ for $p$ and 2 for $b$ in the above formula and simplify for $h$ .

  $\begin{array}{c}{h}^2={\left(-\sqrt{45}\right)}^2+{\left(2\right)}^2\\ h^2=45+4\\ h^2=49\\ h=\pm 7\end{array}$

Since, sides cannot be negative; so, $h=7$

Substitute $-\sqrt{45}$ for $p$ and 7 for $h$ in the formula $\sin u=\frac{p}{h}$ .

  $\sin u=-\frac{\sqrt{45}}{7}$

Substitute 2 for $b$ and 7 for $h$ in the formula $\cos u=\frac{b}{h}$ .

  $\cos u=\frac{2}{7}$

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 $\frac{2}{7}$ for $\cos u$ in the formula $\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\cos u}{2}}$ and simplify.

  $\begin{array}{c}\sin \left(\frac{u}{2}\right)=\sqrt{\frac{1-\frac{2}{7}}{2}}\\ =\sqrt{\frac{\frac{7-2}{7}}{2}}\\ =\sqrt{\frac{5}{14}}\\ \approx 0.59\end{array}$

Substitute $\frac{2}{7}$ for $\cos u$ in the formula $\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\cos u}{2}}$ and simplify.

  $\begin{array}{c}\cos \left(\frac{u}{2}\right)=\sqrt{\frac{1+\frac{2}{7}}{2}}\\ =\sqrt{\frac{\frac{7+2}{7}}{2}}\\ =\sqrt{\frac{9}{14}}\\ \approx 0.8\end{array}$

Substitute $\frac{2}{7}$ for $\cos u$ and $-\frac{\sqrt{45}}{7}$ for $\sin u$ in the formula $\tan \left(\frac{u}{2}\right)=\frac{1-\cos u}{\sin u}$ and simplify.

  $\begin{array}{c}\tan \left(\frac{u}{2}\right)=\frac{1-\frac{2}{7}}{-\frac{\sqrt{45}}{7}}\\ =\frac{\frac{7-2}{\cancel{7}}}{\frac{-\sqrt{45}}{\cancel{7}}}\\ =\frac{5}{-\sqrt{45}}\\ \approx 0.74\end{array}$

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 0.59, 0.8 and 0.74 respectively.