Replace Sin x with 1/nDate: 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/ |
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