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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.

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  • $\begingroup$ 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
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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

$$\sin(2\arccos(Y))$$

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

this gives us

$$2\sin(\arccos(Y))(Y)$$

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

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

so,

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

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

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

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

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

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

$\iff\dfrac\pi4\le2\tan^{-1}x\le\dfrac{3\pi}4$

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

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