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Math Help - Manipulative trigonometry Question?

  1. #1
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    Thumbs down Manipulative trigonometry Question?

    prove that sin10.sin30.sin50.sin70 = 1/16

    well, I grouped together to make this

    (sine10.sin50)(sin30.sin70)

    Since 10 + 50 = 60 , and used the identity sinA.sinB = (cos(a-b)-cos(a+b))/2

    and got the answer in terms of cos40 after further simplification...
    problem is I don't know what cos40 is... Although I could find it out as 40*3 = 120

    and cos(3*40)=cos 120 and then simply using cos3A = 4cos^3(A) - 3cosA
    but then I will have to simplify an annoying cubic equation...please help, do suggest an easier method..
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  2. #2
    Senior Member pacman's Avatar
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    ice_syncer, this method is SIMPLIER.

    Prove: sin10.sin30.sin50.sin70 = 1/16.
    ______________________________________

    sin 30 = cos (90 - 30) = cos 60 = 1/2,

    sin 10 = cos (90 - 10) = cos 80,

    sin 50 = cos (90 - 50) = cos 40,

    sin 70 = cos (90 - 70) = cos 20,

    sin 160 = sin (180 - 160) = sin 20.
    ______________________________________

    Let P = (sin 10 sin 50 sin 70)(sin 30) = (cos 20 cos 40 cos 80)(sin 30)

    multiply BS by sin 20,

    P(sin 20) = (sin 20)(cos 20 cos 40 cos 80)(sin 30)

    use the identity: 2 sin A cos A = sin 2A,

    P(sin 20) = (sin 20 cos 20)(cos 40 cos 80)(sin 30)

    P(sin 20) = (1/2)(sin 40)(cos 40 cos 80)(sin 30)

    P(sin 20) = (1/2)(sin 40cos 40)(cos 80)(sin 30)

    P(sin 20) = (1/2)(1/2)(sin 80)(cos 80)(sin 30)

    P (sin 20) = (1/2)(1/2)(1/2)(sin 160)(sin 30)

    P (sin 20) = (1/2)(1/2)(1/2)(sin 20)(sin 30)

    P (sin 20) = (1/2)(1/2))1/2)(1/2)(sin 20)

    you may cancel sin 20, since sin 20 is not equal to 0.

    P = sin10 sin30 sin50 sin70 = 1/16

    If you multiply by BS by sin 40, it will do the same trick, likewise with sin 80. try it . . . .
    Last edited by pacman; September 14th 2009 at 07:23 AM.
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  3. #3
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    Quote Originally Posted by pacman View Post
    ice_syncer, this method is SIMPLIER.

    Prove: sin10.sin30.sin50.sin70 = 1/16.
    ______________________________________

    sin 30 = cos (90 - 30) = cos 60 = 1/2,

    sin 10 = cos (90 - 10) = cos 80,

    sin 50 = cos (90 - 50) = cos 40,

    sin 70 = cos (90 - 70) = cos 20,

    sin 160 = sin (180 - 160) = sin 20.
    ______________________________________

    Let P = (sin 10 sin 50 sin 70)(sin 30) = (cos 20 cos 40 cos 80)(sin 30)

    multiply BS by sin 20,

    P(sin 20) = (sin 20)(cos 20 cos 40 cos 80)(sin 30)

    use the identity: 2 sin A cos A = sin 2A,

    P(sin 20) = (sin 20 cos 20)(cos 40 cos 80)(sin 30)

    P(sin 20) = (1/2)(sin 40)(cos 40 cos 80)(sin 30)

    P(sin 20) = (1/2)(sin 40cos 40)(cos 80)(sin 30)

    P(sin 20) = (1/2)(1/2)(sin 80)(cos 80)(sin 30)

    P (sin 20) = (1/2)(1/2)(1/2)(sin 160)(sin 30)

    P (sin 20) = (1/2)(1/2)(1/2)(sin 20)(sin 30)

    P (sin 20) = (1/2)(1/2))1/2)(1/2)(sin 20)

    you may cancel sin 20, since sin 20 is not equal to 0.

    P = sin10 sin30 sin50 sin70 = 1/16

    If you multiply by BS by sin 40, it will do the same trick, likewise with sin 80. try it . . . .
    OK next time I'll try to use that trick ie multiplyuing both sides with a constant to obtain a direct relation. Thanks!
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  4. #4
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    Hello ice_syncer
    Quote Originally Posted by ice_syncer View Post
    prove that sin10.sin30.sin50.sin70 = 1/16

    well, I grouped together to make this

    (sine10.sin50)(sin30.sin70)

    Since 10 + 50 = 60 , and used the identity sinA.sinB = (cos(a-b)-cos(a+b))/2

    and got the answer in terms of cos40 after further simplification...
    problem is I don't know what cos40 is... Although I could find it out as 40*3 = 120

    and cos(3*40)=cos 120 and then simply using cos3A = 4cos^3(A) - 3cosA
    but then I will have to simplify an annoying cubic equation...please help, do suggest an easier method..
    Here's an alternative solution, that continues from where you'd got to (and you must have been very close!):

    (\sin10\sin50)(\sin30\sin70)=\tfrac14(\cos40-\cos60)(\cos40-\cos100)

    =\tfrac14(\cos40-\cos60)(\cos40+\cos80)

    =\tfrac14(c-\tfrac12)(c+2c^2-1), where c = \cos40 and \cos80 = 2c^2-1

    =\tfrac18(2c-1)(2c^2+c-1)

    =\tfrac18(4c^3-3c+1)

    =\tfrac18(\cos120+1), using \cos3\theta = 4\cos^3\theta - 3\cos\theta

    =\tfrac{1}{16}

    Grandad
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  5. #5
    Senior Member pacman's Avatar
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    P = sin 10 sin 30 sin 50 sin 70 = 1/16.

    from this book i've got this identity, sin 3x = 4 sin (x) sin (60 - x)sin (60 + x),

    Elementary Trigonometry by M.A., and S. R. Knight, B.A., M.B., CH.B., H. S. Hall (Hardcover - 1955)

    sin 3x = 4 sin (x) sin (60 - x)sin (60 + x), when x = 10 degrees

    we have,

    sin 3(10) = 4 sin 10 sin (60 -10) sin (60 + 10)

    sin 30 = 4 sin 10 sin 50 sin 70, thus

    sin 10 sin 50 sin 70 = (1/4) sin 30 = (1/4)(1/2) = 1/8.

    Then, P = (sin 10 sin 50 sin 70)(sin 30) = (1/8)(1/2) = 1/16

    It does follow that sin 10 sin 30 sin 50 is only a special case of the

    identity: sin 3x = 4 sin (x) sin (60 - x)sin (60 + x), where x = 10 degrees.

    P = (sin 10 sin 50 sin 70)(sin 30) = (1/8)(1/2) = 1/16.
    Last edited by pacman; September 15th 2009 at 12:41 AM.
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  6. #6
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    Thanks!, Never thought of converting cos 100 to -cos80 = -cos(2*40)
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