Inverse Trigonometry System of Equations

Question

$$2\tan^{-1}\left(\sqrt{x-x^2}\right) = \tan^{-1}\left(x\right)\: +\, \tan^{-1}\left(1-x\right)$$

I have a feeling solution includes drawing triangles but cannot make the leap to get the solution

What are you trying to do? Solve that equation for $x$? – Nov 22 '17 at 05:32 — , Nov 22 '17 at 05:32

Deepak · Accepted Answer · 2017-11-22T05:35:11.927

3

Take tangent of both sides (apply double angle and sum of angle formula and remember that $\tan(\tan^{-1}y) = y$):

$$\frac{2\sqrt{x-x^2}}{1 - (x-x^2)} = \frac{x+1-x}{1-x(1-x)}$$

and solve that equation.

edited Nov 22 '17 at 05:35

answered Nov 22 '17 at 05:33

Deepak

How did you get the above equation? Could you elaborate what you did after simply taking the tangent? – Jeffrey Chiu Nov 22 '17 at 05:34
@JeffreyChiu : He used the double-angle formula and the sum of angles formula for tangent. – Eric Towers Nov 22 '17 at 05:35
1

@JeffreyChiu Does my edit help? – Deepak Nov 22 '17 at 05:35
Yes I see now, thanks a lot – Jeffrey Chiu Nov 22 '17 at 05:36
@Deepak, Taking tan can be erroneous, right as $\tan(m\pi+u)=\tan u$ for integer $m$ – lab bhattacharjee Nov 22 '17 at 05:37
@labbhattacharjee Would taking $\frac{2\sqrt{x-x^2}}{1 - (x-x^2)} = \frac{x+1-x}{1-x(1-x)} + k\pi, k \in \mathbb{Z}$ fix this issue? – Deepak Nov 22 '17 at 05:40
@Deepak, It can not be arbitrary $k$. Please check one of the links in my answer. – lab bhattacharjee Nov 22 '17 at 05:42

score 1 · Answer 2 · answered Nov 22 '17 at 05:35

We need $x(1-x)\ge0\iff x(x-1)\le0\iff0\le x\le1$

$\implies\dfrac{x+1-x}2\ge\sqrt{x(1-x)}$

Now use $$\arctan x+\arctan y= \arctan\frac{x+y}{1-xy}$$ if $xy<1$

2 Answers2