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Trig quiz math help

  1. Jan 13, 2005 #1


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    I proposed this problem on a thread in the homework section.I honestly do not know the answer...

    "Compute [tex] \sin\frac{\pi}{5} [/tex] using trigonometry and algebra only."

  2. jcsd
  3. Jan 13, 2005 #2


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    It's actually fairly simple. I haven't worked it all the way out to the bitter end, but I've verified my method and it works.

    Expand [tex]\sin 5\theta[/tex] = [tex]\sin(3\theta + 2\theta)[/tex] in terms of [tex]\sin \theta[/tex]. You would just use the addition formula followed by expanding each of the terms with the triple and double angle formulae.

    All the terms with powers of [tex]\cos\theta[/tex] will have even powers of the cosine, so they can easily be converted to even powers of sine with [tex]\cos^2\theta = 1 - \sin^2\theta[/tex].

    You will eventually get a reducible quintic in terms of sine theta. Let [itex]s = \sin\theta[/itex]

    [tex]\sin 5\theta = 16s^5 - 20s^3 + 5s[/tex]

    Equate that to zero (since [itex]\sin\pi = 0[/itex]) and solve.

    Dismissing [itex]s = 0[/itex], it becomes a quartic which is actually a quadratic in [itex]s^2[/itex], which you can solve to get :

    [tex]s^2 = \frac{1}{8}(5 \pm \sqrt{5})[/tex]

    One of the values (with the plus sign is a redundant root).

    EDIT : I've not yet found a way to determine the explicit value of [itex]s[/itex] from that expression. I keep getting ugly expressions with more roots of surds. I'm still working on this part, but for now, my answer is :

    [tex]\sin{\frac{\pi}{5}} = \sqrt{\frac{1}{8}(5 - \sqrt{5})}[/tex]
    Last edited: Jan 13, 2005
  4. Jan 13, 2005 #3


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    And, as a matter of fact,

    [tex]\cos{\frac{\pi}{5}} = \frac{1}{2}\phi[/tex]

    where [tex]\phi = \frac{1}{2}(1 + \sqrt{5})[/tex], the golden ratio. But the expression for the sine still involves squaring that, subtracting the result from unity, and taking the root, and I cannot find a way to simplify that.
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