How can I compute $\tan(.5\arctan(x))$?

Question

The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.

Answer 1

You have to compute $y=\tan\left( \frac{\arctan{(x)}}{2} \right)$. Let's call $\theta=\arctan(x)$, so you have to compute $y=\tan\left(\frac{\theta}{2}\right)$.

You should write $\tan\left(\frac{\theta}{2}\right)$ in terms of $\tan\left(\theta\right)$, say you use: $$\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\left(\theta\right)}{\sin\left(\theta\right)} = \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}$$ but from $$\cos^2\left(\theta\right)+\sin^2\left(\theta\right)=1$$ we get that $$\frac{1}{\tan^2\left(\theta\right)}+1 = \frac{1}{\sin^2\left(\theta\right)}$$ which is the same of

$$ \frac{1}{\sin\left(\theta\right)}=\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad 0\lt\theta\lt\pi \Rightarrow x \gt 0 \\ $$ $$ \frac{1}{\sin\left(\theta\right)}=-\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad -\pi\lt\theta\lt 0 \Rightarrow x \lt 0\\ $$

so finally we can write

$$\tan\left(\frac{\theta}{2}\right)= \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}=\pm\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}-\frac{1}{\tan\left(\theta\right)}$$

This is very helpfull because $\theta=\arctan(x)$ so $\tan(\arctan(x))=x$ and $$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=+\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\gt 0$$ $$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=-\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\lt 0$$

Answer 2

$$y=\tan(0.5 \arctan(x))$$

$$\arctan(y)= \frac12 \arctan(x)$$

$$x=\tan(2 \arctan(y))$$

using formula of $\tan(2\theta)$

$$x= \frac{2 \tan(\arctan(y))}{1-\tan^2(\arctan(y))}=\frac{2y}{1-y^2}$$

Then you can find $y$.

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For the form of $y=\tan(\beta \arctan(x))$, in general, there is no simplification.

Update

Thanks to Jeevan Devaranjan for his comment mentioning that

$$\tan(\beta \arctan x)=(\frac1i)(\frac{(1+ix)^\beta-(1-ix)^\beta}{(1+ix)^\beta+(1-ix)^\beta})$$

Answer 3

Recall that $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan a \tan b}$. Also let $c = \tan({\frac{\arctan x}{2}})$. \begin{align} c &= \tan(\arctan x - \frac{\arctan x}{2})\\ &= \frac{\tan(\arctan x) - \tan(\frac{\arctan x}{2})}{1 + \tan(\arctan x)\tan(\frac{\arctan}{2})}\\ &= \frac{x - c}{1 + xc}\\ c^2x + c &= x- c\\ c^2x + 2c - x & = 0 \end{align} So by the quadratic formula(note that we take the positive root since due to the domain of the inverse tangent function) \begin{equation} c = \frac{-2 + \sqrt{4 - 4(x)(-x)}}{2x} = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \end{equation} Therefore \begin{equation} \tan(\frac{\arctan x}{2}) = -\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \end{equation}

Answer 4

As shown in this answer, $\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}$. Then, since $\sin\left(\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}}$ and $\cos\left(\tan^{-1}(x)\right)=\frac1{\sqrt{1+x^2}}$, we get $$ \tan\left(\tfrac12\tan^{-1}(x)\right)=\frac{x}{\sqrt{1+x^2}+1} $$ This works for all $x\in\mathbb{R}$.