Please help solve simultaneous inverse trig equations

In summary, the conversation discusses a pair of inverse trigonometric equations involving arctan and arccos functions. The equations are derived from a puzzle problem involving two ladders and a passageway. The conversation presents different approaches to solving the equations, including using complex math and treating the angles as angles in a triangle. The final solution involves a quartic equation and the use of properties of logarithms, resulting in the values of x and y being approximately 11.074 and 5.138, respectively.
  • #1
eldrick
52
0
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 :confused: )
 
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  • #2
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.
 
  • #3
are you sure you can go further from that ?
 
  • #4
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? :\
 
  • #5
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.
 
  • #6
dav2008 said:
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 :confused:
 
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  • #7
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 )
 
  • #8
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: [tex]0 \leq \arccos (x) \leq \pi[/tex], so that means:
[tex]\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) \geq 0[/tex], hence x > 0, since [tex]\arctan \left( \frac{8}{x} \right) \geq 0[/tex].
Do the same for the second equation to obtain y > 0.
Let:
[tex]\arctan \left( \frac{8}{x} \right) = \arccos \left( \frac{x + y}{20} \right) = \alpha[/tex]. That means:
[tex]\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}[/tex]
And from [tex]\arccos \left( \frac{x + y}{20} \right) = \alpha[/tex], you'll have: [tex]\cos \alpha = \frac{x + y}{20}[/tex], so we have:
[tex]\frac{(x + y) ^ 2}{400} = \frac{x ^ 2}{x ^ 2 + 64} \quad (1)[/tex].
Do the same for the second equation to obtain:
[tex]\frac{(x + y) ^ 2}{900} = \frac{y ^ 2}{y ^ 2 + 64} \quad (2)[/tex]
Eleminate (x + y)2 from 2 equations above (Equation (1), and equation (2)) to abtain:
[tex]\frac{x ^ 2}{x ^ 2 + 64} = \frac{9}{4} \times \frac{y ^ 2}{y ^ 2 + 64}[/tex]
Solve for y in terms of x, we have:
[tex]y ^ 2 = \frac{256 x ^ 2}{5 x ^ 2 + 576} \quad (3)[/tex]
Substitute (3) into (1), we have:
[tex]\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}[/tex]
[tex]\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[/tex]
[tex]\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)[/tex]
[tex]\Leftrightarrow ( 5 x ^ 2 + 576 + 32 \sqrt{5x ^ 2 + 576} + 256 ) (x ^ 2 + 64) = 400 (5 x ^ 2 + 576)[/tex]
[tex]\Leftrightarrow (32 \sqrt{5x ^ 2 + 576}(x ^ 2 + 64) = 400 (5 x ^ 2 + 576) - (5x ^ 2 + 832) (x ^ 2 + 64)[/tex]
From there, square both sides, you'll have a quartic equation in x2 :yuck:
Something like this:
[tex]25 x^8-13600 x^6-2297600 x^4+203980800 x^2+28966912000 = 0[/tex]
And there are 2 real roots:
[tex]\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.[/tex]
The second x is not valid since [tex]\frac{x + y}{20} \approx \frac{25.6988 + y}{20} > 1[/tex]
From the first x, we have:
[tex]y = \sqrt {\frac{256 x ^ 2}{5 x ^ 2 + 576}} = \frac{16 x}{\sqrt{5x ^ 2 + 576}} \approx 5.138[/tex]
So the solution is:
[tex]\left\{ \begin{array}{l} x \approx 11.074 \\ y \approx 5.138 \end{array} \right.[/tex]
 
Last edited:

What are simultaneous inverse trig equations?

Simultaneous inverse trig equations are a set of two or more equations that involve inverse trigonometric functions such as sin^-1, cos^-1, tan^-1, etc. These equations are solved simultaneously to find the values of the variables.

Why are simultaneous inverse trig equations important?

Simultaneous inverse trig equations are important in many real-world applications, especially in physics, engineering, and navigation. They are used to find unknown angles or distances in a triangle or other geometric shapes.

What are the steps to solve simultaneous inverse trig equations?

The steps to solve simultaneous inverse trig equations are:

  1. Isolate the inverse trig functions on one side of the equation.
  2. Use the inverse function to cancel out the trig function.
  3. Apply the appropriate trigonometric identities to simplify the equation.
  4. Solve for one variable at a time.
  5. Check the solution by substituting the values back into the original equations.

Can simultaneous inverse trig equations have multiple solutions?

Yes, simultaneous inverse trig equations can have multiple solutions. This is because inverse trig functions have more than one possible input for a given output. It is important to check the solutions and determine which ones satisfy all the equations in the system.

What are some common mistakes made when solving simultaneous inverse trig equations?

Some common mistakes made when solving simultaneous inverse trig equations include:

  • Forgetting to apply the inverse function to cancel out the trig function.
  • Not using the appropriate trigonometric identities to simplify the equations.
  • Incorrectly solving for a variable, leading to wrong solutions.
  • Not checking the solutions by substituting them back into the original equations.

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