# Question b5a12

Oct 19, 2017

see below

#### Explanation:

$\sin x = \frac{1}{9}$. We need the ratio for cos x so we can calculate the $\left(\frac{x}{2}\right)$. So since we know that the opposite is 1 and the hypotenuse is 9 and we are in quadrant 1 from the given domain we can calculate the adjacent side using pythagorean theorem.

That is,
$a = \sqrt{{9}^{2} - {1}^{2}} = \sqrt{81 - 1} = \sqrt{80} = 4 \sqrt{5}$. Hence,

$\cos x = \frac{4 \sqrt{5}}{9} , 0 < x < \frac{\pi}{2}$. We also need the domain for$\frac{x}{2}$ and we can get that by halving the given domain $0 < \frac{x}{2} < \frac{\pi}{4}$ so $\frac{x}{2}$ is in Quadrant I. Since it is in quadrant I $\sin \left(\frac{x}{2}\right) , \cos \left(\frac{x}{2}\right) \mathmr{and} \tan \left(\frac{x}{2}\right)$ will all have positive answers.

Now lets calculate $\sin \left(\frac{x}{2}\right) , \cos \left(\frac{x}{2}\right) \mathmr{and} \tan \left(\frac{x}{2}\right)$,

sin (x/2)=sqrt(1/2 (1-cosx)

=sqrt(1/2 (1-(4sqrt5)/9)

=sqrt((9-4sqrt5)/18

$= \frac{\sqrt{162 - 72 \sqrt{5}}}{18}$

$\cos \left(\frac{x}{2}\right) = \sqrt{\frac{1}{2} \left(1 + \cos x\right)}$

=sqrt(1/2(1+(4sqrt5)/9)

=sqrt((9+4sqrt5)/18

$= \frac{\sqrt{162 + 72 \sqrt{5}}}{18}$

$\tan \left(\frac{x}{2}\right) = \sin \frac{x}{1 + \cos x}$

=(1/9)/(1+(4sqrt5)/9

$= \frac{\frac{1}{9}}{\frac{9 + 4 \sqrt{5}}{9}}$

$= \frac{1}{\cancel{9}} \cdot \frac{\cancel{9}}{9 + 4 \sqrt{5}}$

$= \frac{1}{9 + 4 \sqrt{5}} \cdot \frac{9 - 4 \sqrt{5}}{9 - 4 \sqrt{5}}$

=(9-4sqrt5)/(81-80#

$= 9 - 4 \sqrt{5}$