# How to simplify elegantly $\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)$?

I currently try to simplify the following trigonometric expression: $$\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)$$ where $t\in(0;1]$.

I know that $\arctan(x)=\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$ and I am also aware that there are formulas to simplify $\arcsin(x)+\arcsin(y)$, but they depend on different cases, as it can be seen here.

Is there a rather elegant way to solve the problem?

Any help is highly appreciated.

## 4 Answers

Set $$f(t):=\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right), \qquad t \in (0,1]$$ by differentiating $$f'(t)=0,\qquad t \in (0,1]$$ thus $f$ is constant on $(0,1]$, putting $t=1$ you get $f(1)=\dfrac\pi2$ giving

$$\arcsin(2t-1)+2\arctan\left(\sqrt{\frac{1-t}{t}}\right)=\frac\pi2 \qquad t \in (0,1]$$

Let $\arctan\sqrt{\dfrac{1-t}t}=u$

Though the principal value of $\arctan$ lies in $\in\left[-\dfrac\pi2,\dfrac\pi2\right]$

As $\sqrt{\dfrac{1-t}t}\ge0\implies0\le u\le\dfrac\pi2\iff0\le2u\le\pi$

and $\dfrac{1-t}t=\tan^2u\implies t=\cos^2u$

$\implies\arcsin(2t-1)=\arcsin(2\cos^2u-1)=\arcsin\cos2u=\dfrac\pi2-\arccos(\cos2u)$

Now $\arccos(\cos2u)=\begin{cases} 2u &\mbox{if } 0\le2u\le\pi \\ -2u & \mbox{if } -\pi\le2u<0\end{cases}$

Hope you can take it from here.

$\displaystyle \sin^{-1}(2t-1)+2\tan^{-1}\left(\sqrt{\frac{1-t}{t}}\right)\;,$ Where $0<t\leq 1$

Now Using $\displaystyle \sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}\Rightarrow \sin^{-1}(x) = \frac{\pi}{2}-\cos^{-1}(x)$

Now expression is $\displaystyle \frac{\pi}{2}-\cos^{-1}(2t-1)+2\tan^{-1}\left(\sqrt{\frac{1-t}{t}}\right)$

Now Put $t=\cos^2 \phi\;,$ Then $0<\cos^2 \phi \leq 1\Rightarrow -1 \leq \cos \phi \leq 1-\left\{0\right\}$

So we get $\displaystyle = \frac{\pi}{2}-\cos^{-1}\left(2\cos^2 \phi -1 \right)+2\tan^{-1}\left(\sqrt{\frac{1-\cos^2 \phi}{\sin^2 \phi}}\right)$

So we get $\displaystyle = \frac{\pi}{2}-\cos^{-1}(\cos 2\phi)+2\tan^{-1}\left(|\tan \phi|\right)\;,$ Where $\displaystyle 0 \leq \phi \leq \pi-\left\{\frac{\pi}{2}\right\}$

Here is a nice and fun (maybe elegant?) approach. For $$\frac{1}{2} consider the Figure below.

$$\hskip1.5in$$

$$\triangle ABC$$ is a right-angled triangle with side $$\overline{BC} = 2t-1$$ and hypotenuse $$\overline{AC} = 1$$, so that $$\alpha = \arcsin(2t-1).$$ By Pythagorean Theorem, $$\overline{AB} = 2\sqrt{t-t^2}.$$ Extend $$BC$$ to a segment $$\overline{BD} = 2t$$, so that $$\overline{CD} = 1$$ and $$\beta = \arctan\sqrt{\frac{1-t}{t}}.$$ Using the fact that $$\triangle ACD$$ is isosceles and $$\triangle ABD$$ is right-angled we can write $$(\beta + \alpha) + \beta = \frac{\pi}{2},$$ i.e. $$$$\arcsin(2t-1) + 2\arctan\sqrt{\frac{1-t}{t}} = \frac{\pi}{2}.\tag{1}\label{1}$$$$

For $$0 consider the following Figure. $$\hskip1.5in$$

Now $$\overline{BC} = -2t+1$$, and $$\overline{AC} = 1$$, so that $$\alpha = -\arcsin(2t-1).$$ Again we have $$\overline{AB} = 2\sqrt{t-t^2}$$. Extend $$BC$$ to a segment $$\overline{CD} = 1$$, so that $$\overline{BD} = 2$$ and $$\beta = \arctan\sqrt{\frac{1-t}{t}}.$$ We obtain again \eqref{1} by considering realtionships between $$\alpha$$ and $$\beta$$ due to the fact that $$\triangle ACD$$ is isosceles, i.e. $$\beta +(\beta - \alpha) = \frac{\pi}{2}.$$ $$\blacksquare$$