# 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?

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 $$\sin x = -\sqrt{\dfrac {1-3/5}2} = -\dfrac {1}{\sqrt 5}$$.

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.

• Wolfram is wrong.....GASP. Commented Jan 29, 2021 at 12:49
• @A-LevelStudent Surprising, isn’t it Commented Jan 29, 2021 at 15:21
• Exactly, agreed. Commented Jan 29, 2021 at 15:22

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, $$\sin\theta=\frac{-1}{\sqrt{5}}$$

$$\pm\sqrt{[1+\cos \left( \cot^{-1(...)})]/2\right)}$$
$$\pm \sqrt{\dfrac{1\pm\frac35 }{2}}$$
$$\pm \dfrac{2}{\sqrt 5}, \pm\dfrac{1}{\sqrt 5}.$$