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I have verified with Mathematica that, for $R>0, y \in \mathbb{R}$:

$$ \tan{\frac{\arcsin{\frac{y}{R}}}{2}} = \frac{R - \sqrt{R^2 -y ^2}}{y}$$

using 

Assuming[Element[y, Reals] && R > 0, 
 FullSimplify[TrigToExp[Tan[ArcSin[y/R]/2]]]]

How can I prove this withouse messing with complex exponentials and logarithms?

Nathaniel Bubis
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José D.
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2 Answers2

2

Create a right triangle with sides $x,y$, and hypotenuse $R$. In the diagram below, $\phi = \arcsin(y/R) / 2$, So you are looking to find $\tan \phi$ in terms of $R$ and $y$.

For this purpose, you may find the angle bisector theorem useful, as it tells you how the two sections of $y$ are related: $$\tan \phi = \frac{y_1}{x}=\frac{y_2}{R}$$ Or that: $$y_1 = \left(\frac{R}{x}+1\right)^{-1}y \quad\longrightarrow\quad \tan \phi=\frac{y}{\sqrt{R^2-y^2}+R}$$ Now multiply the denominator by it's conjugate to obtain your result.

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

Nathaniel Bubis
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1

Let $\displaystyle\arcsin\frac yr=\phi\implies \sin\phi=\frac yR\ \ \ \ (1)$

and based on the definition of principal value, $\displaystyle-\frac\pi2\le\phi\le\frac\pi2\ \ \ \ (2)$

So, we need to find $\displaystyle\tan\frac\phi2$

Using Weierstrass substitution, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}$

So, using $\displaystyle(1),\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac yR\implies y\tan^2\frac\phi2-2R\tan\frac\phi2+y=0\ \ \ \ (3)$

Solving the Quadratic Equation for $\displaystyle\tan\frac\phi2=\frac{R\pm\sqrt{R^2-y^2}}y$

Now using $\displaystyle(2), -\frac\pi4\le\frac\phi2\le\frac\pi4\implies -1\le\tan\frac\phi2\le1$

Observe that for real $\displaystyle\phi, R^2\ge y^2\implies R\ge |y|$ as $R>0$

If $\displaystyle R=|y|,\tan\frac\phi2=\pm1$ where both roots are same

Else $\displaystyle\frac{R+\sqrt{R^2-y^2}}{|y|}> \frac R{|y|}>1,$ hence should be discarded.

Observe that if $\displaystyle\tan\frac{\phi_1}2,\tan\frac{\phi_2}2$ are the roots of $(3),$

using Vieta's formula, $\displaystyle\tan\frac{\phi_1}2\tan\frac{\phi_2}2=1$

$\displaystyle\implies\tan\frac{\phi_1}2=\frac1{\tan\frac{\phi_2}2}=\cot\frac{\phi_2}2=\tan\left(\frac\pi2-\frac{\phi_2}2\right)=\tan\frac{\pi-\phi_2}2$

$\displaystyle\implies\frac{\phi_1}2=\frac{\pi-\phi_2}2\iff\phi_1=\pi-\phi_2 $

$\displaystyle\implies\sin\phi_1=\sin\left(\pi-\phi_2\right)=\sin\phi_2 $

Also, $\displaystyle\sin\phi=\frac{2\tan\frac\phi2}{1+\tan^2\frac\phi2}=\frac{2\cot\frac\phi2}{1+\cot^2\frac\phi2}$ (multiplying the numerator & the denominator by $\displaystyle\cot^2\frac\phi2$)

lab bhattacharjee
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