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Math Help - Cos and Tan

  1. #1
    Super Member dhiab's Avatar
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    Cos and Tan

    Prove : \cos 36^\circ > \tan 36^\circ <br />
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  2. #2
    MHF Contributor red_dog's Avatar
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    \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}

    \sin 36^{\circ}=\frac{(\sqrt{5}-1)\sqrt{1+2\sqrt{5}}}{8}

    Then \tan 36^{\circ}=\frac{(3-\sqrt{5})\sqrt{10+2\sqrt{5}}}{4}

    We have to prove \sqrt{5}+1>(3-\sqrt{5})\sqrt{10+2\sqrt{5}}

    Square both members:

    6+2\sqrt{5}>(14-6\sqrt{5})(10+2\sqrt{5})

    17\sqrt{5}>37

    Square again both members: 1440>1369 which is true.
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  3. #3
    Senior Member pacman's Avatar
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    cos A = tan A

    cos A - sin A / cos A = 0

    cos^2 A - sin A = 0

    (1 - sin^2 A) - sin A = 0

    sin^2 A + sin A - 1 = 0

    sin A = (-1 +/- (1 - 4(1)(11))^(1/2)/[2(1)]

    sin A = (-1 +/- sqrt 5)/2

    A = 45 degress = pi/4 rad

    that means if A = 0 to A<pi/4, cos A > tan A
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