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Math Help - Cos 3pi/5

  1. #1
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    Cos 3pi/5

    I don't understand the step where 2Sin^2(3pi/10) = 2((sqrt5+1)/4) Cos 3pi/5-steps.jpg
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  2. #2
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    Re: Cos 3pi/5

    Hey Eraser147.

    Try setting up a right angled triangle or using sums or differences formula to get the value.
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  3. #3
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    Re: Cos 3pi/5

    Consider the right-angled triangle with acute angles  \frac{3\pi}{10} and  \frac{2\pi}{10} and note that \sin{\left(\frac{3\pi}{10}\right)}=\cos{ \left( \frac{2\pi}{10}\right) }.

    To ease typing, let \frac{\pi}{10}=A, then \sin 3A=\cos 2A.

    Use the standard identities, \sin 3A =3\sin A-4\sin^{3} A .............(1) \text{  and  } \cos 2A= 1-2\sin^{2} A ............(2)

    and we have the cubic 4\sin^{3} A -2\sin^{2}A-3\sin A +1 = 0.

    (\sin A -1) is a factor, remove this and we are left with the quadratic

    4\sin^{2} A +2\sin A -1=0, and solving this, (for the angle in the 0,\pi/2 range),

    \sin A = \frac{-1+\sqrt{5}}{4}.

    Substituting this into (1) and simplifying leads to \sin 3A = \sin{ \ \left( \frac{3\pi}{10} \right)} = \frac{1+\sqrt{5}}{4}.
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