# Represent inverse trigonometric equation as rational integral equation?

• Apr 28th 2009, 12:29 AM
fardeen_gen
Represent inverse trigonometric equation as rational integral equation?
Prove that the equation:
$\displaystyle \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$
can be represented as a rational integral equation in x & y.

NOTE: No latex for arccot?
EDIT: @Mush:
Code:

\acot
doesn't work!
EDIT: Thanks Soroban! Also, $\displaystyle \text{arccot}$ is the only way it seems.
• Apr 28th 2009, 02:00 AM
Mush
Quote:

Originally Posted by fardeen_gen
Prove that the equation:
$\displaystyle \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$
can be represented as a rational integral equation in x & y.

NOTE: No latex for arccot?

\acot
• Apr 28th 2009, 06:42 AM
Soroban
Hello, fardeen_gen!

Quote:

Prove that the equation: .$\displaystyle \text{arccot}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) \;=\; 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$

can be represented as a rational integral equation in $\displaystyle x$ and $\displaystyle y.$

$\displaystyle \text{We have: }\;\arctan\underbrace{\left(\frac{\sqrt{1 - x^2 - y^2}}{y}\right)}_{\theta} \:= \:2\cdot\underbrace{\arctan\sqrt{\frac{3 - 4x^2}{4x^2}}}_{\alpha} - \underbrace{\arctan\sqrt{\frac{3 - 4x^2}{x^2}}}_{\beta}$

. . That is: .$\displaystyle \theta \;=\;2\alpha - \beta$

Take the tangent of both sides: .$\displaystyle \tan\theta \;=\;\tan(2\alpha + \beta)$

. . $\displaystyle \tan\theta \;=\;\frac{\tan(2\alpha) + \tan\beta} {1 - \tan(2\alpha)\tan\beta} \;=\; \frac{\dfrac{2\tan\alpha}{1 - \tan^2\!\alpha} + \tan\beta}{1 - \dfrac{2\tan\alpha}{1 - \tan^2\!\alpha}\tan B}$

. . $\displaystyle \tan\theta \;=\;\frac{2\tan\alpha - \tan\beta(1 - \tan^2\!\alpha)} {1-\tan^2\!\alpha - 2\tan\alpha\tan\beta}$

. . $\displaystyle \tan\theta - \tan^2\!\alpha\tan\theta - 2\tan\alpha\tan\beta\tan\theta \;=\;2\tan\alpha - \tan\beta + \tan^2\!\alpha\tan\beta$ .[1]

Since $\displaystyle \tan\theta = \frac{\sqrt{1-x^2-y^2}}{y},\quad \tan\alpha = \frac{\sqrt{3-4x^2}}{2x},\quad \tan\beta = \frac{\sqrt{3-4x^2}}{x}$

substitute into [1]:

. . $\displaystyle \frac{\sqrt{1-x^2-y^2}}{y} \;-\; \frac{3-4x^2}{4x^2} \cdot \frac{\sqrt{1-x^2-y^2}}{y} \;-\; 2 \cdot \frac{\sqrt{3-4x^2}}{2x}\cdot\frac{\sqrt{3-4x^2}}{x}\cdot\frac{\sqrt{1-x^2-y^2}}{y}$

. . . . $\displaystyle = \;2\,\frac{\sqrt{3-4x^2}}{2x} - \frac{\sqrt{3-4x^2}}{x} + \frac{3-4x^2}{4x^2}\cdot\frac{\sqrt{3-4x^2}}{x}$

Simplify: .$\displaystyle \frac{\sqrt{1-x^2-y^2}}{y} - \frac{5(3-4x^2)\sqrt{1-x^2-y^2}}{4x^2y} \;=\;\frac{(3-4x^2)^{\frac{3}{2}}}{4x^3}$

Multiply by $\displaystyle 4x^3y\!:\quad 4x^3\sqrt{1-x^2-y^2} - 5x(3-4x^2)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}}$

Factor and simplify: .$\displaystyle 3x(8x^2-5)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}}$

Square both sides: .$\displaystyle 9x^2(8x^2-5)^2(1-x^2-y^2) \;=\;y^2(3-4x^2)^3$

. . and we have a polynomial equaton in $\displaystyle x$ and $\displaystyle y.$