Replace Sin x with 1/n

Date: 12/19/2001 at 13:30:28
From: Eric White
Subject: Trigonometry

(5 + 2sqrt(6))^(sin x) + (5 - 2sqrt(6))^(sin x) = 2sqrt(3)

Date: 12/19/2001 at 16:58:32
From: Doctor Jaffee
Subject: Re: Trigonometry

Hi Eric,

Here is one approach you can take. I successfully solved the problem 
using this method. Sin x has to be a number between 1 and -1, 
inclusive, so let's replace sin x with 1/n.

The problem now says 

   (5 + 2sqrt(6))^(1/n) + (5 - 2sqrt(6))^(1/n) = 12^(1/2)

If you raise both sides to the n power and use the binomial theorem, 
you will get an expansion of n + 1 terms on the left side and 12^(n/2) 
on the right.

What I planned to do next was trial an error.  Let n = 2,3,4,5, etc.  
However, I discovered right away that n = 2 worked.

Let's take another look at it.  Let a = 5 + 2sqrt(6) and 
b = 5 - 2sqrt(6). Let sin(x) = 1/n

The problem now says a^(1/n) + b^(1/n) = 12^(1/2). Now, let's guess 
that n = 2 and square both sides. If that doesn't work, we'll assume 
that n = 3, then cube both sides, and if that doesn't work, we'll let 
n = 4, etc...

But n = 2 works.   
  [ a^(1/2) + b^(1/2)]^2 = [12^(1/2)]^2.
a + 2a^(1/2)*b^(1/2) + b = 12
     a + 2(ab)^(1/2) + b = 12

Replace a with 5 + 2sqrt(6) and b with 5 - 2sqrt(6) and we get

5 + 2sqrt(6) + 2[(5 + 2sqrt(6))(5 - 2sqrt(6))]^(1/2) + 5 - 2sqrt(6) 
                      = 12
10 + 2(25 - 24)^(1/2) = 12
                   12 = 12

Since sin x = 1/n and n = 2, we have sin x = 1/2, so, assuming that 
-90 <= x <= 90, x = 30 degrees.

This solution may be a little unsatisfactory since it depends somewhat 
on trial and error, but it works, and I don't believe there is an 
easier way to do it.

Write back if you want to discuss this probem some more.

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/