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Math Help - Trigonometric identity

  1. #1
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    Trigonometric identity

    Can somebody solve this and give detailed description ?
    I get stuck on this part 1/2 sin18..
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  2. #2
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    sin 18 degrees

    Hello ldx2

    Welcome to Math Help Forum!
    Quote Originally Posted by ldx2 View Post
    Can somebody solve this and give detailed description ?
    I get stuck on this part 1/2 sin18..
    I agree: \sin^224^o-\sin^26^o = \tfrac12\sin 18^o.

    There are various ways of deriving \sin18^o. Here's one. And here's my version:
    \sin 18^o = \cos 72^o = 2\cos^236^o-1=2(1-2\sin^218)^2-1
    So if we put  x = \sin18^o, x is the root of the equation:
    x = 2(1-2x^2)^2-1
    for which 0<x<1.
    \Rightarrow x=2(1-4x^2+4x^4)-1

    \Rightarrow 8x^4-8x^2-x+1=0
    Noting that x = 1 satisfies this equation, we can factorise:
    (x-1)(8x^3+8x^2-1)=0
    and then discard the factor (x-1), since 0<x<1.

    We then note that x = -\tfrac12 is also a root. Hence (2x+1) is a factor, which again we discard for the same reason; leaving us with the remaining factor:
    4x^2+2x-1=0
    the positive root of which is:
    x = \frac{\sqrt5-1}{4}=\sin18^o
    Hence \sin^224^o-\sin^26^o = \frac{\sqrt5-1}{8}

    Grandad
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  3. #3
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    Thanks for the quick reply and solving it. EDIT: I understood until the part where you say x=-1/2 is also a root and how you found out (2x+1) is also a factor. Could you explain that part in more detail ?
    Last edited by ldx2; December 14th 2009 at 10:13 AM.
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  4. #4
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    Hello ldx2
    Quote Originally Posted by ldx2 View Post
    Thanks for the quick reply and solving it. EDIT: I understood until the part where you say x=-1/2 is also a root and how you found out (2x+1) is also a factor. Could you explain that part in more detail ?
    By inspection (which is a posh word for 'trial and error'), when x = -\tfrac12, 8x^3+8x^2-1= -1+2-1=0. So, using the remainder theorem, (2x+1) is a factor.

    Grandad
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  5. #5
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    LOL 'trial and error' ! Thank you for clearing it up. Could you please take a look at my other thread where i posted 2 trig. problems !
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