# a in(pi,(3pi)/2);sina=-4/5;How to calculate tga/2?

Apr 3, 2017

$\tan \left(\frac{a}{2}\right) = \frac{1}{2}$ or $\tan \left(\frac{a}{2}\right) = - 2$

#### Explanation:

Since $a \setminus \in \left(\setminus \pi , \frac{3 \setminus \pi}{2}\right)$, the angle $a$ is in Quadrant III, where the $x$ (length of the adjacent side) and $y$ (length of the opposite side) are both negative. Denote $z$ as the length of the hypotenuse. ($z$ is always positive.)

We have that $\sin a = - \frac{4}{5} \setminus \Leftrightarrow \frac{y}{z} = \frac{- 4}{5}$.

Note that the ratio of $y$ to $z$ is $- 4$ to $5$, but we don't necessarily have $y = - 4$ and $z = 5$. However, we can assume so because we work with ratios ($\sin , \cos , \tan$ of angles) and not exact lengths of a triangle.

By the Pythagorean theorem,
${x}^{2} + {y}^{2} = {z}^{2}$
$\setminus R i g h t a r r o w {x}^{2} + {\left(- 4\right)}^{2} = {5}^{2}$
$\setminus R i g h t a r r o w {x}^{2} + 16 = 25$
$\setminus R i g h t a r r o w {x}^{2} = 9$
$\setminus R i g h t a r r o w x = - 3$ (refer to first line, $x$ must be negative in Quadrant III)

So $\tan a = \frac{y}{x} = \frac{- 4}{- 3} = \frac{4}{3}$.

Now use the double-angle identity $\tan \left(2 x\right) = \frac{2 \tan x}{1 - {\tan}^{2} x}$.
For $x = \frac{a}{2}$, the identity becomes

tan a=(2tan(a/2))/(1-tan^2(a/2)
\Rightarrow 4/3=(2tan(a/2))/(1-tan^2(a/2)
\Rightarrow 2/3=(tan(a/2))/(1-tan^2(a/2)
$\setminus R i g h t a r r o w 2 - 2 {\tan}^{2} \left(\frac{a}{2}\right) = 3 \tan \left(\frac{a}{2}\right)$
$\setminus R i g h t a r r o w 0 = 2 {\tan}^{2} \left(\frac{a}{2}\right) + 3 \tan \left(\frac{a}{2}\right) - 2$
$\setminus R i g h t a r r o w 0 = \left(2 \tan \left(\frac{a}{2}\right) - 1\right) \left(\tan \left(\frac{a}{2}\right) + 2\right)$
$\setminus R i g h t a r r o w \left(2 \tan \left(\frac{a}{2}\right) - 1\right) = 0$ or $\tan \left(\frac{a}{2}\right) + 2 = 0$
$\setminus R i g h t a r r o w \tan \left(\frac{a}{2}\right) = \frac{1}{2}$ or $\tan \left(\frac{a}{2}\right) = - 2$.