Help finding trig values with the given information I need help with the following problem:

Find $\sin x/2$, $\cos x/2$, and $\tan x/2$ from the given information.
$\cot x = 4,   180^\circ < x < 270^\circ$

I thought $\sin x/2$ might be:

$\sqrt{1-(4/\sqrt{17})/2}$

But that was wrong.
I was able to get $\tan x/2$:

$-(4+\sqrt{17})$

Any help would be appreciated.
Thanks
 A: $\cot(x)=\frac{\cos(x)}{\sin(x)}=\frac{\cos(x/2)^2-\sin(x/2)^2}{2\sin(x/2)\cos(x/2)}$
and observe that $\sin(x/2)^2+\cos(x/2)^2=1$
A: I've confirmed that the value you computed for $\tan (x/2)=-4-\sqrt{17}$ is right. Since $180{{}^\circ}<x<270{{}^\circ}$, hence $90^\circ <x/2 <135^\circ $, it's the negative solution of the equation
$$\tan \left( x\right) =\frac{2\tan \left(
x/2\right) }{1-\tan ^{2}\left( x/2\right) }=\frac{1}{4}.\tag {1}
$$
It can be used to evaluate e.g. $\sin (x/2)$ as follows. Express $\sin^2 (x/2) $ as a function of $\tan^2 (x/2) $, by dividing both sides of the fundamental identity
\begin{equation*}
\sin ^{2}(x/2)+\cos ^{2}(x/2)=1\tag {2}
\end{equation*}
by $\sin ^{2}(x/2)$
$$1+\cot^2 (x/2)=\frac {1}{\sin^2 (x/2)}; \tag {3}$$
and, since $\cot (x/2)=1/\tan (x/2)$, rewrite $(3) $ as
\begin{equation*}
\sin ^{2}(x/2)=\frac{\tan ^{2}(x/2)}{1+\tan ^{2}(x/2)}.\tag {4}
\end{equation*}
To compute $\sin (x/2)$ take into account that for $90^\circ <x/2 <135^\circ $ we have $\sin (x/2)>0$. 
Finally compute $\cos (x/2)$.
