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Math Help - Trig

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Trig

    Show  \displaystyle \tan \frac\pi9 + 4\sin \frac\pi9 = \sqrt3 .
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  2. #2
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    Quote Originally Posted by chiph588@ View Post
    Show  \displaystyle \tan \frac\pi9 + 4\sin \frac\pi9 = \sqrt3 .

    Consider  \frac{\pi}{3} - \frac{\pi}{9} = \frac{2\pi}{9}


     \sin(\frac{\pi}{3} - \frac{\pi}{9}) = \sin(\frac{2\pi}{9})

     \frac{\sqrt{3}}{{2}} \cos(\frac{\pi}{9})  - \frac{1}{2} \sin(\frac{\pi}{9})  = 2\sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})

     \sqrt{3} \cos(\frac{\pi}{9})  - \sin(\frac{\pi}{9})  = 4 \sin(\frac{\pi}{9})  \cos(\frac{\pi}{9})

     \sqrt{3} = \tan(\frac{\pi}{9})  + 4 \sin(\frac{\pi}{9})
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by simplependulum View Post
    Consider  \frac{\pi}{3} - \frac{\pi}{9} = \frac{2\pi}{9}


     \sin(\frac{\pi}{3} - \frac{\pi}{9}) = \sin(\frac{2\pi}{9})

     \frac{\sqrt{3}}{{2}} \cos(\frac{\pi}{9})  - \frac{1}{2} \sin(\frac{\pi}{9})  = 2\sin(\frac{\pi}{9}) \cos(\frac{\pi}{9})

     \sqrt{3} \cos(\frac{\pi}{9})  - \sin(\frac{\pi}{9})  = 4 \sin(\frac{\pi}{9})  \cos(\frac{\pi}{9})

     \sqrt{3} = \tan(\frac{\pi}{9})  + 4 \sin(\frac{\pi}{9})
    Much more elegant than my solution!
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  4. #4
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    Quote Originally Posted by chiph588@ View Post
    Much more elegant than my solution!
    What was your solution? De Moivre/binomial/Vieta?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    What was your solution? De Moivre/binomial/Vieta?
    Let  u = \cos\left(\frac\pi9\right) . From the third angle formula, we obtain

     8u^3-6u-1 = 0

     -(2u+1)(8u^3-6u-1) = 0

     -16u^4-8u^3+12u^2+8u+1 = 0

     (1-u^2)(1+4u)^2 = 3u^2

     \sqrt{1-u^2}(1+4u) = \sqrt3u

     \sqrt{1-u^2}+4u\sqrt{1-u^2} = \sqrt3u

     \sin\left(\frac\pi9\right)+4\cos\left(\frac\pi9\ri  ght)\sin\left(\frac\pi9\right) = \sqrt3\cos\left(\frac\pi9\right)

     \tan\left(\frac\pi9\right)+4\sin\left(\frac\pi9\ri  ght) = \sqrt3
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Let  u = \cos\left(\frac\pi9\right) . From the third angle formula, we obtain

     8u^3-6u-1 = 0

     -(2u+1)(8u^3-6u-1) = 0

     -16u^4-8u^3+12u^2+8u+1 = 0

     (1-u^2)(1+4u)^2 = 3u^2

     \sqrt{1-u^2}(1+4u) = \sqrt3u

     \sqrt{1-u^2}+4u\sqrt{1-u^2} = \sqrt3u

     \sin\left(\frac\pi9\right)+4\cos\left(\frac\pi9\ri  ght)\sin\left(\frac\pi9\right) = \sqrt3\cos\left(\frac\pi9\right)

     \tan\left(\frac\pi9\right)+4\sin\left(\frac\pi9\ri  ght) = \sqrt3
    If you're wondering how I came up with this, it's called working backwards .
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