# How do you find tan x/2 ; given sin x= 3/5, with 90<x<180?

Jun 6, 2015

There is a property of the $\tan$ function that states:

if $\tan \left(\frac{x}{2}\right) = t$ then
$\sin \left(x\right) = \frac{2 t}{1 + {t}^{2}}$

From here you write the equation

$\frac{2 t}{1 + {t}^{2}} = \frac{3}{5}$

$\rightarrow 5 \cdot 2 t = 3 \left(1 + {t}^{2}\right)$

$\rightarrow 10 t = 3 {t}^{2} + 3$

$\rightarrow 3 {t}^{2} - 10 t + 3 = 0$

Now you find the roots of this equation:

$\Delta = {\left(- 10\right)}^{2} - 4 \cdot 3 \cdot 3 = 100 - 36 = 64$

${t}_{-} = \frac{10 - \sqrt{64}}{6} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$

${t}_{+} = \frac{10 + \sqrt{64}}{6} = \frac{10 + 8}{6} = \frac{18}{6} = 3$

Finaly you have to find which of the above answers is the right one. Here is how you do it:

Knowing that 90°< x <180° then 45°< x/2 <90°

Knowing that on this domain, $\cos \left(x\right)$ is a decreasing function and $\sin \left(x\right)$ is an increasing function, and that sin(45°) = cos(45°)
then $\sin \left(\frac{x}{2}\right) > \cos \left(\frac{x}{2}\right)$

Knowing that $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ then in our case $\tan \left(\frac{x}{2}\right) > 1$

Therefore, the correct answer is $\tan \left(\frac{x}{2}\right) = 3$