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?
 A: Since $$0\lt \cot^{-1} (-\frac 34 ) \lt \pi$$
, $$0\lt \frac 12 \cot^{-1} (-\frac 34) \lt \frac{\pi}{2} $$ which means its sine must be positive, so Wolfram’s answer is wrong. $\frac{2}{\sqrt 5}$ is correct.
A: Both are correct, but the answer you obtain depends on the definition $\cot^{-1}(x)$, in particular its range, since inverse trigonometric functions are multi-valued.
If you write $\cot^{-1}(x) = \tan^{-1}(1/x)$ [as a calculator seldom has $\cot^{-1}$] and use the range $-\pi/2 < \tan^{-1}x < \pi/2$, we see that $\theta = \cot^{-1}(-3/4)$ must be negative.
In $-\pi/2 < x < 0$, $\cos x > 0$ and $\sin x < 0$. Hence $\cos \theta = \dfrac35$, and $\sin x = -\sqrt{\dfrac {1-3/5}2} = -\dfrac {1}{\sqrt 5}$.
A: You got the expression $\cot \theta= \frac{-3}{4}$
As you may know, the value of $\cot \theta$ is postive in the first and third quadrants, which means it is negative in the second and fourth quadrants.
When $\cot \theta=\frac{-3}{4}$, $\cos \theta$ can either equal $\frac{-3}{5}$ or $\frac{3}{5}$
$\cos\theta=\frac{3}{5}$ is true wrt the fourth quadrant.
If $\cos \theta=\frac{3}{5}$, $\sin \theta= \frac{1}{\sqrt{5}}$
But, we know that in the fourth quadrant $\sin \theta$ is negative.
Thus, $\sin\theta=\frac{-1}{\sqrt{5}}$
In short, both answers are correct, for their respective quadrants.
A: 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.
