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Math Help - Evaluate sin18

  1. #1
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    Evaluate sin18

    I need to calculate the exact value of sin18. How do I do that? I see that \frac {90}{5}=18, so I must use some form of the addition theorems. What do I do?

    BTW, how do I type the degree symbol in latex?
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  2. #2
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    Quote Originally Posted by chengbin View Post
    I need to calculate the exact value of sin18. How do I do that? I see that \frac {90}{5}=18, so I must use some form of the addition theorems. What do I do?

    BTW, how do I type the degree symbol in latex?
    Read this: Trigonometry--The derivation of Sin 18&#176

    [tex]18^0[/tex] is easiest.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Read this: Trigonometry--The derivation of Sin 18

    [tex]18^0[/tex] is easiest.
    That's an interesting page (particularly the derivation of \sin 1^0.

    So if you spent enough time, you can find the exact value of all the integer degrees of \sin and \cos from 0 to 180?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by chengbin View Post
    I need to calculate the exact value of sin18. How do I do that? I see that \frac {90}{5}=18, so I must use some form of the addition theorems. What do I do?

    BTW, how do I type the degree symbol in latex?
    see this link Trigonometry Angles--Pi/5 -- from Wolfram MathWorld

    in the link above you can find \cos 36^o
    but

    \cos 2x = 1-2\sin ^2 x so


    \cos 36^o = 1-2\sin ^2 18^o
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    [tex]18^0[/tex] is easiest.
    But [tex]18^\circ[/tex] is better!

    18^0\qquad18^\circ
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  6. #6
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    Quote Originally Posted by chengbin View Post
    I need to calculate the exact value of sin18. How do I do that? I see that \frac {90}{5}=18, so I must use some form of the addition theorems. What do I do?

    BTW, how do I type the degree symbol in latex?
    Hello This the solution :


    \begin{array}{l}<br />
let:x = 18^\circ \\ <br />
5x = 90^\circ \\ <br />
3x + 2x = 90^\circ \\ <br />
3x = 90^\circ - 2x \\ <br />
\sin 3x = \cos 2x \\ <br />
\end{array}
    I'have this equation :
    \begin{array}{l}<br />
3\sin x - 4\sin ^3 x = 1 - 2\sin ^2 x \\ <br />
let:\sin x = y \\ <br />
4y^3 - 2y^2 - 3y + 1 = 0 \\ <br />
\left( {y - 1} \right)\left( {4y^2 + 2y - 1} \right) = 0 \\ <br />
\end{array}<br />
    I'deduct:
    \begin{array}{l}<br />
y_1 = 1...NO \\ <br />
y_2 = \frac{{ - 1 - \sqrt 5 }}{2} < 0...NO \\ <br />
y_3 = \frac{{ - 1 + \sqrt 5 }}{2} = \sin 18^\circ \\ <br />
\end{array}<br />
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  7. #7
    Senior Member pacman's Avatar
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    sin 18 = cos 72. we have solved this already at http://www.mathhelpforum.com/math-he...tml#post344761

    Working out in degrees, we have
    Let x =72 degrees, multiply both sides by 5;
    5x = 360
    2x + 3x = 360
    2x = 360-3x

    Then, taking cos of both side
    cos 2x = cos(360 -3x)
    cos 2x = cos 3x

    Using an identity for double angle and triple angle, equating them we have
    2cos^2 x -1= 4cos^3 x - 3 cos x
    4cos^3 x - 2cos^2 x - 3 cos x +1=0
    4cos^3 x - 4cos^2 x + 2 cos^2 x - 2 cos x - cos x +1=0
    4cos^2x(cos x -1) +2cos x( cos x -1) -1( cos x - 1) =0
    (cos x - 1) (4cos^x +2 cos x - 1) =0

    Either cos x =0 means x=90 degree
    Solve (4cos^x +2 cos x - 1) =0

    cos x = [-2 +(plus or minus){sq rt (4+16)] /8
    cos x = [-2 +(plus or minus){sq rt (20)] /8
    cos x = [-2 +2(plus or minus){sq rt (5)] /8
    cos x = [-1 +(plus or minus){sq rt (5)] /4

    Neglecting NEGATIVE values of cos x, because x lies in first quadrant
    Therefore cos 72 degrees = (sqrt(5) -1)/4

    cos 72 = 0.30901699437494742410229341718 . . . . = sin 18
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  8. #8
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    Quote Originally Posted by chengbin View Post
    I need to calculate the exact value of sin18. How do I do that? I see that \frac {90}{5}=18, so I must use some form of the addition theorems. What do I do?

    BTW, how do I type the degree symbol in latex?
    Check this post.

    http://www.mathhelpforum.com/math-he...tml#post350000

    \sin54^\circ-\sin18^\circ=\cos36^\circ-\sin18^\circ=1-2\sin^2 18^\circ-\sin18^\circ=\frac{1}{2}

    It is a quadratic equation.
    Last edited by mr fantastic; September 18th 2009 at 10:04 AM. Reason: Restored original reply
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  9. #9
    Senior Member pacman's Avatar
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    luobo, nice insight. here is the graph of your equation . . . .
    Attached Thumbnails Attached Thumbnails Evaluate sin18-luobo.gif  
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  10. #10
    Senior Member pacman's Avatar
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    dhiab, nice solution . . . . here is the graph of your equation, 4y^3 - 2y^2 - 3y + 1 = 0.
    Attached Thumbnails Attached Thumbnails Evaluate sin18-cubic.gif  
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