The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.
So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?

In all four values. First, half angle formula
$$ \pm\sqrt{[1+\cos \left( \cot^{-1(...)})]/2\right)}$$
$$\pm \sqrt{\dfrac{1\pm\frac35 }{2}}$$
$$\pm \dfrac{2}{\sqrt 5}, \pm\dfrac{1}{\sqrt 5}.$$
So both answers correct.
