# Given Cotx=5, how do you find sin (x/2), cos (x/2), tan (x/2)?

Feb 13, 2017

$\sin \left(\frac{x}{2}\right) = \frac{- 5 + \sqrt{26}}{\sqrt{52 - 10 \sqrt{26}}}$ $\cos \left(\frac{x}{2}\right) = \frac{1}{\sqrt{52 - 10 \sqrt{26}}}$ and $\tan \left(\frac{x}{2}\right) = - 5 + \sqrt{26}$

or $\sin \left(\frac{x}{2}\right) = \frac{- 5 - \sqrt{26}}{\sqrt{52 + 10 \sqrt{26}}}$, $\cos \left(\frac{x}{2}\right) = \frac{1}{\sqrt{52 + 10 \sqrt{26}}}$ and $\tan \left(\frac{x}{2}\right) = - 5 - \sqrt{26}$

#### Explanation:

As $\cot x = 5$, $\tan x = \frac{1}{5}$

but as $\tan x = \frac{2 \tan \left(\frac{x}{2}\right)}{1 - {\tan}^{2} \left(\frac{x}{2}\right)}$

$\frac{2 \tan \left(\frac{x}{2}\right)}{1 - {\tan}^{2} \left(\frac{x}{2}\right)} = \frac{1}{5}$

or $10 \tan \left(\frac{x}{2}\right) = 1 - {\tan}^{2} \left(\frac{x}{2}\right)$

or ${\tan}^{2} \left(\frac{x}{2}\right) + 10 \tan \left(\frac{x}{2}\right) - 1 = 0$

$\tan \left(\frac{x}{2}\right) = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \times 1 \times \left(- 1\right)}}{2}$

= $\frac{- 10 \pm \sqrt{104}}{2} = - 5 \pm \sqrt{26}$

i.e. $\tan \left(\frac{x}{2}\right) = - 5 + \sqrt{26}$ or $- 5 - \sqrt{26}$

Hence $\sec \left(\frac{x}{2}\right) = \sqrt{1 + {\left(- 5 + \sqrt{26}\right)}^{2}}$

= $\sqrt{1 + 25 + 26 - 10 \sqrt{26}} = \sqrt{52 - 10 \sqrt{26}}$ or

$\sec \left(\frac{x}{2}\right) = \sqrt{1 + {\left(- 5 - \sqrt{26}\right)}^{2}}$

= $\sqrt{1 + 25 + 26 + 10 \sqrt{26}} = \sqrt{52 + 10 \sqrt{26}}$

and $\cos \left(\frac{x}{2}\right) = \frac{1}{\sqrt{52 - 10 \sqrt{26}}}$ or $\frac{1}{\sqrt{52 + 10 \sqrt{26}}}$

and $\sin \left(\frac{x}{2}\right) = \frac{- 5 + \sqrt{26}}{\sqrt{52 - 10 \sqrt{26}}}$ or $\frac{- 5 - \sqrt{26}}{\sqrt{52 + 10 \sqrt{26}}}$