# How do you find the value?

## $\cos$ $\frac{\alpha}{2}$ if $\tan \alpha = \frac{40}{9} , \left(180 < \alpha < 270\right)$

Feb 19, 2018

$\cos \left(\frac{\alpha}{2}\right) = - \frac{4}{\sqrt{41}}$

#### Explanation:

Given:
$\left(180 < \alpha < 270\right)$
indicating that $\cos \alpha$ is negative

$\tan \alpha = \frac{40}{9}$
opposite side is 40
hypotenuse will be 41,
because
$\sqrt{{9}^{2} + {40}^{2}} = \sqrt{81 + 1600} = \sqrt{1681} = 41$
$\cos \alpha = \text{(adjacent side)/(hypotenuse)}$
$\cos \alpha = - \frac{9}{41}$
cos(alpha/2)=+-sqrt((1+cosalpha)/2)=+-sqrt((1-9/41)/2
$= \pm \sqrt{\frac{41 - 9}{41 \times 2}} = \pm \sqrt{\frac{32}{2} \times \frac{1}{41}} = \pm \frac{\sqrt{16}}{\sqrt{41}}$

$\cos \left(\frac{\alpha}{2}\right) = \pm \frac{4}{\sqrt{41}}$

As mentioned, ${180}^{\circ} < \alpha < {270}^{\circ} ,$
it follows that
$\frac{180}{2} < \frac{\alpha}{2} < \frac{270}{2}$

$90 < \frac{\alpha}{2} < 135$, where $\cos \left(\frac{\alpha}{2}\right)$ is negative

Hence,
$\cos \left(\frac{\alpha}{2}\right) = - \frac{4}{\sqrt{41}}$

Feb 19, 2018

${\tan}^{2} \left(\alpha\right) + 1 = {\sec}^{2} \left(\alpha\right)$

Substitute ${\sec}^{2} \left(\alpha\right) = \frac{1}{\cos} ^ 2 \left(\alpha\right)$

${\tan}^{2} \left(\alpha\right) + 1 = \frac{1}{\cos} ^ 2 \left(\alpha\right)$

Multiply both sides by ${\cos}^{2} \frac{\alpha}{{\tan}^{2} \left(\alpha\right) + 1}$:

${\cos}^{2} \left(\alpha\right) = \frac{1}{{\tan}^{2} \left(\alpha\right) + 1}$

Use the square root operation on both sides:

$\cos \left(\alpha\right) = \pm \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}$

We are told that ${180}^{\circ} < \alpha < {270}^{\circ}$, therefore we choose the negative value:

$\cos \left(\alpha\right) = - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}$

$1 + \cos \left(\alpha\right) = 1 - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}$

Multiply both sides by $\frac{1}{2}$:

$\frac{1 + \cos \left(\alpha\right)}{2} = \frac{1 - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}}{2}$

Use the square root operation on both sides:

$\pm \sqrt{\frac{1 + \cos \left(\alpha\right)}{2}} = \pm \sqrt{\frac{1 - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}}{2}}$

Substitute $\pm \sqrt{\frac{1 + \cos \left(\alpha\right)}{2}} = \cos \left(\frac{\alpha}{2}\right)$

$\cos \left(\frac{\alpha}{2}\right) = \pm \sqrt{\frac{1 - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}}{2}}$

From ${180}^{\circ} < \alpha < {270}^{\circ}$ we derive ${90}^{\circ} < \frac{\alpha}{2} < {135}^{\circ}$ and conclude that the cosine function is negative within the specified domain:

$\cos \left(\frac{\alpha}{2}\right) = - \sqrt{\frac{1 - \sqrt{\frac{1}{{\tan}^{2} \left(\alpha\right) + 1}}}{2}}$

Substitute ${\tan}^{2} \left(\alpha\right) = {\left(\frac{40}{9}\right)}^{2}$:

$\cos \left(\frac{\alpha}{2}\right) = - \sqrt{\frac{1 - \sqrt{\frac{1}{{\left(\frac{40}{9}\right)}^{2} + 1}}}{2}}$

I used WolframAlpha to simplify the above into an exact form:

$\cos \left(\frac{\alpha}{2}\right) = - \frac{4 \sqrt{41}}{41}$