Iterative scheme to arrive at a solution?

Jeff Byram · **Joined:** Tue, 21 Apr 2009 21:10:30 UTC **Posts:** 1

Good readers, I wish to find a value for Y given any value of X for the equation:

acos(Y/((X^2+Y^2)^.5))+atan((Y-X)/(Y+X))-pi/4=0

How can this be done? I have tried and failed to use a ten year old Hewlett Pakard 48SX in equation solving mode to solve for Y given any value of X.

Are there any other calculators that can do this? Can someone recommend a course of action to solve this equation? Can the solution be graphed?

Am I correct in assuming that it can only be solved by some kind of iterative scheme?

Please advise!

simplethinker · **Joined:** Fri, 9 Nov 2007 00:04:38 UTC **Posts:** 128

Jeff Byram wrote:

Good readers, I wish to find a value for Y given any value of X for the equation:

acos(Y/((X^2+Y^2)^.5))+atan((Y-X)/(Y+X))-pi/4=0

How can this be done? I have tried and failed to use a ten year old Hewlett Pakard 48SX in equation solving mode to solve for Y given any value of X.

Are there any other calculators that can do this? Can someone recommend a course of action to solve this equation? Can the solution be graphed?

Am I correct in assuming that it can only be solved by some kind of iterative scheme?

Please advise!

Try $y=\log_x(1)$

Your equation holds identically (it's true for all X and Y).
We have $\arccos \left( \frac{y}{\sqrt{x^2+y^2}} \right) + \arctan \left( \frac{y-x}{y+x} \right) - \frac{\pi}{4} = 0$

Moving the pi/4 to the other side and taking the tangent of the equation yields $\tan \left( \arccos \left( \frac{y}{\sqrt{x^2+y^2}} \right) + \arctan \left( \frac{y-x}{y+x} \right) \right) = \tan(\frac{\pi}{4})$

The addition formula for tangent is $\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}$ . Remembering that trig functions are nothing but ratios: if $\phi = \arccos\left( \frac{y}{\sqrt{x^2 + y^2}} \right)$ , then $\cos(\phi) = \frac{y}{\sqrt{x^2 + y^2}}$ and $\sin(\phi) = \sqrt{1 - \cos(\phi)^2} = \frac{x}{\sqrt{x^2 + y^2}}$ so $\tan(\phi) = \frac{y}{x} = \tan\left(\arccos\left( \frac{y}{\sqrt{x^2 + y^2}} \right)\right)$ .

Now plugging this back into your equation yields $$ \frac{ \frac{y}{x} + \frac{y-x}{y+x} }{1 - \frac{y}{x} \frac{y - x}{y + x}} = \frac{x(y+x) + y(y-x)}{y(y+x) - x(y-x)} = \frac{x^2 + y^2}{y^2 + x^2} = 1$ , so we finally get 1=1.

S.O.S. Mathematics CyberBoard

Iterative scheme to arrive at a solution?

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