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Math Help - numerical identity

  1. #1
    Senior Member pacman's Avatar
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    numerical identity

    prove that cos 36 - sin 18 = 1/2
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  2. #2
    ynj
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  3. #3
    Senior Member pacman's Avatar
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    ynj, thanks.

    i did not know that it has a geometric solution, i failed to scan the forum.

    i do it this way, though very long.

    prove that cos 36 - sin 18 = 1/2. I will evaluate the left hand side of the equation; since cos 36 = sin 54, then

    sin 54 - sin 18 = sin (36 + 18) - sin 18; using the addition law for sine

    [sin (A + B) = (sin A)(cos B) - (cos A)(sin B);

    (sin 36)(cos 18) - (cos 36)(sin 18) - sin 18 = ?

    Ah, i will use the double-angle identity

    sin 2A = 2(sin A)(cos A) and cos 2A = 1 - 2 sin^2 A.

    And also cos^2 A = 1 - sin^2 A. We have,

    2(sin 18)(cos 18)(cos 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18)(cos^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18)(1 - sin^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18) - 2(sin^3 18) + sin 18 - 2(sin^3 18) - sin 18 = ?

    2(sin 18) - 2(sin^3 18) - 2(sin^3 18) = ?

    2(sin 18) - 4(sin^3 18) = ?

    from http://www.mathhelpforum.com/math-he...tml#post350585

    sin 18 = (5^(1/2) - 1)/4, substituting to the above expression and

    expanding it [2(sin 18) - 4(sin^3 18)];

    2[(5^(1/2) - 1)/4] - 4[(5^(1/2) - 1)/4]^3 = ?

    [(5^(1/2) - 1)/2] - 4[(5^(3/2)- 3(5)(1) + 3(5)^(1/2) - 1)/4^3] = ?

    (5^(1/2) - 1)/2 - [5(5^(1/2) - 15 + 3(5)^(1/2) - 1)/4^2] = ?

    (5^(1/2) - 1)/2 - [8(5^(1/2) - 16)/16] = ?

    [(5^(1/2)]/2 - 1/2 - [(5^(1/2)]/2 + 1 = ?

    -1/2 + 1 = ?

    Thus, cos 36 - sin 18 = 1/2.
    Last edited by pacman; August 24th 2009 at 03:35 AM.
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  4. #4
    ynj
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    Quote Originally Posted by pacman View Post
    ynj, thanks.

    i did not know that it has a geometric solution, i failed to scan the forum.

    i do it this way, though very long.

    prove that cos 36 - sin 18 = 1/2. I will evaluate the left hand side of the equation; since cos 36 = sin 54, then

    sin 54 - sin 18 = sin (36 + 18) + sin 18; using the addition law for sine
    [sin (A + B) = (sin A)(cos B) - (cos A)(sin B);

    (sin 36)(cos 18) - (cos 36)(sin 18) - sin 18 = ?

    Ah, i will use the double-angle identity

    sin 2A = 2(sin A)(cos A) and cos 2A = 1 - 2 sin^2 A.

    And also cos^2 A = 1 - sin^2 A.


    We have,

    2(sin 18)(cos 18)(cos 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18)(cos^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18)(1 - sin^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?

    2(sin 18) - 2(sin^3 18) + sin 18 - 2(sin^3 18) - sin 18 = ?

    2(sin 18) - 2(sin^3 18) - 2(sin^3 18) = ?

    2(sin 18) - 4(sin^3 18) = ?

    from http://www.mathhelpforum.com/math-he...tml#post350585

    sin 18 = (5^(1/2) - 1)/4, substituting to the above expression and

    expanding it [2(sin 18) - 4(sin^3 18)];

    2[(5^(1/2) - 1)/4] - 4[(5^(1/2) - 1)/4]^3 = ?

    [(5^(1/2) - 1)/2] - 4[(5^(3/2)- 3(5)(1) + 3(5)^(1/2) - 1)/4^3] = ?

    (5^(1/2) - 1)/2 - [5(5^(1/2) - 15 + 3(5)^(1/2) - 1)/4^2] = ?

    (5^(1/2) - 1)/2 - [8(5^(1/2) - 16)/16] = ?

    [(5^(1/2)]/2 - 1/2 - [(5^(1/2)]/2 + 1 = ?

    -1/2 + 1 = ?

    Thus, cos 36 - sin 18 = 1/2.
    Actually, we do have a geometric solution!! But I don't know how to post picture in the forum.
    I try to state it in a abstract way(if you are really interested in it ,try to draw the graph according to my statement..)
    Let \triangle ABC be a triangle, \angle ABC=\angle ACB=72^{\circ}, \angle BAC=36^{\circ}, D\in AB,\angle DCB=36^{\circ},BC=1,AB=AC=x,we connect C,D.
    It is easy to know \triangle ABC~ \triangle CDB,AD=CD=BC. Thus \frac{AB}{BC}=\frac{BC}{BD}\Longleftrightarrow x=\frac{1}{BD}, so BD=\frac{1}{x}. So AD=AB-BD=x-\frac{1}{x}=BC=1.Thus x-\frac{1}{x}=1.
    Solve this equation, we get x=\frac{1+\sqrt{5}}{2}. But it is also obvious that \frac{\frac{1}{2}BC}{AB}=\cos 72^{\circ}=\sin 18^{\circ}, so we can get \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}!!!
    Then all the things can be solved..
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  5. #5
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    prove that cos 36 - sin 18 = 1/2
    since cos2A=1-sin^2(A)

    cos36=1-2sin^2(18)

    Therefore,

    1-2sin^2(18)-sin(18)
    =1-2((5)^{1/2} - 1)^2/16 - ((5)^{1/2}-1)/4
    =1-(5-2(5)^{1/2}+1)/8 - ((5)^{1/2})/4 + 1/4
    =1/2

    use the fact that sin18={sqrt(5)-1}/4
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  6. #6
    Senior Member pacman's Avatar
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    bandedkrait, your solution is concise and simplier than mine. Thanks
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