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Math Help - I dont even know how to describe that... :S finding an exact value of cos36??

  1. #1
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    I dont even know how to describe that... :S finding an exact value of cos36??

    This one says: Noting that t=pi/5 satisfies 3t=pi-2t , fins the exact value of cos36.....This identities may be uself: cos^2t+sin^2t=1 , sin2t=2sintcost sin3t=3sint-4sin^3t Any clues???
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  2. #2
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    Re: I dont even know how to describe that... :S finding an exact value of cos36??

    Finding the exact value of \cos 36 is the same as finding the same value of \cos \frac{\pi}{5}=\cos t, because \frac{180}{5}=36.

    You are given formulae for sin, so we're going to use them:

    3t=\pi-2t\Rightarrow  \sin 3t=\sin (\pi-2t)=\sin 2t=2\sin t \cos t

    But we can also do this:

    \sin 3t=3 \sin t - 4\sin^3 t

    Since \sin 3t is the same no matter how we write it:

    2\sin t \cos t=3 \sin t - 4\sin^3 t


    t \in \left ( 0, \frac{\pi}{2} \right )\Rightarrow \sin t\neq 0, that means we can simplify the relation by \sin t.

    2 \cos t=3  - 4\sin^2 t \Rightarrow 2\cos t=3-4(1-cos^2 t)

    \Rightarrow 2\cos t=3-4+4\cos^2 t \Rightarrow 4 \cos t^2-2\cos t-1=0


    Let x be \cos t, so x>0.

    4 \cos t^2-2\cos t-1=0 \Leftrightarrow 4x^2-2x-1=0

    \Delta =(-2)^2-4\cdot 4\cdot (-1)=4+16=20


    x_1=\frac{2+\sqrt{20}}{8}=\frac{2+2\sqrt{5}}{8}= \frac{1+\sqrt{5}}{4}>0

    x_2=\frac{2-\sqrt{20}}{8}=\frac{2-2\sqrt{5}}{8}=\frac{1-\sqrt{5}}{4} < 0


    So x=\cos t =\frac{1+\sqrt{5}}{4}.


    PS: I think this thread should be here.
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