Please help solve simultaneous inverse trig equations

[eldrick]

A friend of mine solved these pair of inverse trig equations simultaneously, claiming he did it with high school maths.

Well, I have a similar background ( & am a keen puzzle solver ) & could get nowhere with them !

Here they are :

ArcTan (8/x) = ArcCos ((x+y)/20)

ArcTan (8/y) = ArcCos ((x+y)/30)

Solve these inverse trigonometric equations for x & y

( I will be extrememely embarrased if the method is trivial, so please forgive me if I've wasted your time )

 

[Pseudo Statistic]

ArcTan (8/x) = ArcCos ((x+y)/20)

ArcTan (8/y) = ArcCos ((x+y)/30)

OK... here goes.
arctan x = i/2 log ((1-ix)/(1+ix))
And..
arc cos x = -i log(x + sqrt(x^2 - 1))
So, going back to the first equation...

i/2 log ((1 - 8i/x)/(1 + 8i/x)) = -i log ( (x+y)/20 + sqrt ((x+y)^2/20^2 - 1)
And the second equation...
i/2 log ((1 - 8i/y)/(1 + 8i/y)) = -i log ( (x+y)/30 + sqrt ((x+y)^2/30^2 - 1)

The Is cancel out, and using the properties of logarithms (namingly if log x = log y, x = y and k log x = log (x^k), you'll end up with a nonlinear system of two equations to solve... doesn't look like it'll be very pretty. ;)

BTW, there's probably some relation between arctan and arccos or something if your friend could do it in high school-- either that or he/she's taken (or somehow learnt) complex math in school.

 

[Hamhiu]

are you sure you can go further from that ?

 

[Pseudo Statistic]

Yup:
i/2 log ((1 - 8i/x)/(1 + 8i/x)) = -i log ( (x+y)/20 + sqrt ((x+y)^2/20^2 - 1)
And the second equation...
i/2 log ((1 - 8i/y)/(1 + 8i/y)) = -i log ( (x+y)/30 + sqrt ((x+y)^2/30^2 - 1)
Cancelling the Is...

1/2 log ((1 - 8i/x)/(1 + 8i/x)) = -1 log ( (x+y)/20 + sqrt ((x+y)^2/20^2 - 1)
And the second equation...
1/2 log ((1 - 8i/y)/(1 + 8i/y)) = -1 log ( (x+y)/30 + sqrt ((x+y)^2/30^2 - 1)

Property of logarithms-> k log x = log(x^k):
log [[(1 - 8i/x)/(1 + 8i/x)]^1/2] = log ( [(x+y)/20 + sqrt ((x+y)^2/20^2 - 1)]^-1 )
And the second equation...
log ([(1 - 8i/y)/(1 + 8i/y)]^1/2) = log ([ (x+y)/30 + sqrt ((x+y)^2/30^2 - 1)]^-1 )

Property log x = log y -> x = y:
[(1 - 8i/x)/(1 + 8i/x)]^1/2 = [(x+y)/20 + sqrt ((x+y)^2/20^2 - 1)]^-1
And the second equation...
[(1 - 8i/y)/(1 + 8i/y)]^1/2 = [(x+y)/30 + sqrt ((x+y)^2/30^2 - 1)]^-1

And from there (writing it down on a piece of paper makes it look a whole lot easier BTW) either use a numerical method (e.g. Newton-Raphson) to solve it or whatever seems appropriate...
It might even reduce to a linear equation after a lot of simplification, who knows? :\

 

[dav2008]

I tried the approach of treating the angle in each equation as an angle in a triangle.

That didn't get me very far though. I just got x=0=y.

 

[eldrick]

I tried the approach of treating the angle in each equation as an angle in a triangle.

That didn't get me very far though. I just got x=0=y.

This was the method i tried & arc cos ((x + y )) 20, implies a triangle with sides : ( x + y ) , ( (400 - ( x + y )^2 ) ^ 1/2 ) , 20

from which you can get an arc tan & equate it to 8 / x

( i did the same for the 2nd equation )

You therefore end up with an equations with no trig, but polynomials in x, y & ( x + y )

Unfortunately this reduced down to a quartic

 

[eldrick]

I'll give the context of those equations :

It comes from a classic ladder problem in a puzzle book ( it's among the toughest of 500 puzzles ) :

https://www.amazon.com/gp/product/0486443418/?tag=pfamazon01-20

it's problem number 400 & states :

"2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

A friend of mine did a diagram, from which the original equations were derived - i hope this link works :

http://i2.tinypic.com/x3h1mv.jpg

anyhows, try an method you like ( the above is just one method of attack )

 

[VietDao29]

Uhmm, I doubt that your friend is telling the truth. This problem looks horrible.
Now arccos(x) is always equal to or greater than 0, right?, since: 0 \leq \arccos (x) \leq \pi, so that means:
\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) \geq 0, hence x > 0, since \arctan \left( \frac{8}{x} \right) \geq 0.
Do the same for the second equation to obtain y > 0.
Let:
\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) = \alpha. That means:
\tan \alpha = \frac{8}{x} \Leftrightarrow 1 + \tan ^ 2 \alpha = \frac{x ^ 2 + 64}{x ^ 2} \Leftrightarrow \frac{1}{\cos ^ 2 \alpha} = \frac{x ^ 2 + 64}{x ^ 2} \Leftrightarrow \cos ^ 2 \alpha = \frac{x ^ 2}{x ^ 2 + 64}
And from \arccos \left( \frac{x + y}{20} \right) = \alpha, you'll have: \cos \alpha = \frac{x + y}{20}, so we have:
\frac{(x + y) ^ 2}{400} = \frac{x ^ 2}{x ^ 2 + 64} \quad (1).
Do the same for the second equation to obtain:
\frac{(x + y) ^ 2}{900} = \frac{y ^ 2}{y ^ 2 + 64} \quad (2)
Eleminate (x + y)² from 2 equations above (Equation (1), and equation (2)) to abtain:
\frac{x ^ 2}{x ^ 2 + 64} = \frac{9}{4} \times \frac{y ^ 2}{y ^ 2 + 64}
Solve for y in terms of x, we have:
y ^ 2 = \frac{256 x ^ 2}{5 x ^ 2 + 576} \quad (3)
Substitute (3) into (1), we have:
\frac{x ^ 2 + \frac{32 x ^ 2}{\sqrt{5x ^ 2 + 576}} + \frac{256 x ^ 2}{5 x ^ 2 + 576}}{400} = \frac{x ^ 2}{x ^ 2 + 64}
\Leftrightarrow \left( x ^ 2 + \frac{32 x ^ 2}{\sqrt{5x ^ 2 + 576}} + \frac{256 x ^ 2}{5 x ^ 2 + 576} \right) (x ^ 2 + 64) = 400 x ^ 2
\Leftrightarrow ( x ^ 2 (5 x ^ 2 + 576) + 32 x ^ 2 \sqrt{5x ^ 2 + 576} + 256 x ^ 2 ) (x ^ 2 + 64) = 400 x ^ 2 (5 x ^ 2 + 576)
\Leftrightarrow ( 5 x ^ 2 + 576 + 32 \sqrt{5x ^ 2 + 576} + 256 ) (x ^ 2 + 64) = 400 (5 x ^ 2 + 576)
\Leftrightarrow (32 \sqrt{5x ^ 2 + 576}(x ^ 2 + 64) = 400 (5 x ^ 2 + 576) - (5x ^ 2 + 832) (x ^ 2 + 64)
From there, square both sides, you'll have a quartic equation in x²
Something like this:
25 x^8-13600 x^6-2297600 x^4+203980800 x^2+28966912000 = 0
And there are 2 real roots:
\left[ \begin{array}{l} x ^ 2 \approx 122.6338468 \\ x ^ 2 \approx 660.42895867809 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x \approx 11.074 \\ x \approx 25.6988 \end{array} \right.
The second x is not valid since \frac{x + y}{20} \approx \frac{25.6988 + y}{20} > 1
From the first x, we have:
y = \sqrt {\frac{256 x ^ 2}{5 x ^ 2 + 576}} = \frac{16 x}{\sqrt{5x ^ 2 + 576}} \approx 5.138
So the solution is:
\left\{ \begin{array}{l} x \approx 11.074 \\ y \approx 5.138 \end{array} \right.