Convert trigonometric function into algebraic function. $$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$

My approach is as follows:

$sin2\theta$ is to be calculated

$$\tan^{-1}x=\gamma \tag{1}$$

Hence $$\begin{align} \cos\theta&=\cot2\gamma \tag{2}\\[4pt] \cos\theta&=\frac{\cot^2\gamma-1}{2\cot\gamma} \tag{3}\\[4pt] \cos\theta&=\frac{1-x^2}{2x} \tag{4} \end{align}$$

The value of $\sin\theta$ is coming in negative.

  • $\begingroup$ check your formula for $\cos \theta$ $\endgroup$ – Vasya May 4 at 13:14
  • $\begingroup$ You have introduced the symbol $\theta$ without telling us what it stands for. $\endgroup$ – Gerry Myerson May 4 at 13:20

I firstly apologise about how this is written, I really ought to learn how to type this stuff properly. Secondly, if there's any mistakes please do say!

$$\tan(2x) = 2\tan(x)/1-\tan^2(x)$$

so $$\cot(2x) = (1-\tan^2(x))/2\tan(x)$$

swapping $x = \arctan(x)$

$$\cot(2\arctan(x)) = (1-x^2)/2x$$

Okay, that bit is done,

Let $(1-x^2)/2x = Y$ for simplicity...

So we want to evaluate


Well, $$\sin(2x)= 2\sin(x)\cos(x)$$

this gives us


So we just want to now evaluate $$2\sin(\arccos(Y))$$

Well $$\sin^2(x)+\cos^2(x)=1$$


$$\sin(x) = \sqrt {(1 - \cos^2(x)})$$

$$\sin(\arccos(y)) = \sqrt {(1-Y^2)}$$

$$2\sin(\arccos(y)) = \sqrt{2(1-Y^2)}$$

So the answer is $$2Y \sqrt {(1-Y^2)}$$

where $Y=(1-x^2)/2x$.


If $\cos^{-1}y=u,\cos u=y,0\le y\le\pi$

If $\sin(2\cos^{-1}y)=\sin2u=2\sin u\cos u=\begin{cases}2u\sqrt{1-u^2} &\mbox{if }0\le y\le\dfrac\pi2 \\ -2u\sqrt{1-u^2} & \mbox{if } y>\dfrac\pi2 \end{cases} $

Here $u=\dfrac{1-x^2}{2x}$

For real $\cos^{-1}y,-1\le\cot(2\tan^{-1}x)\le1$


$\iff\tan\dfrac\pi8\le x\le\tan\dfrac{3\pi}8$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.