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Math Help - Trignometric Question

  1. #1
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    Arbroath, Bonnie Scotland.
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    Trignometric Question

    Hi,
    Can you help me with the following question please?

    3sinx = sin3x

    My working:
    3sinx - sin3x = 0
    3sinx - sin(x + 2x) = 0
    3 sinx - sinx - sin2x = 0
    2sinx - 2sinxcosx =0
    2sinx ( 1 - cos x) = 0

    2 sinx = 0 x= 0 x = 0 and 360 degrees
    1 - cos x = 0 cos x = 1 x = 0 and 360 degrees.

    Can you help I think I have missed a value between 0 and 360?

    Thanks
    Cromlix
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  2. #2
    Junior Member
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    Quote Originally Posted by cromlix View Post
    Hi,
    Can you help me with the following question please?

    3sinx = sin3x

    My working:
    3sinx - sin3x = 0
    3sinx - sin(x + 2x) = 0
    3 sinx - sinx - sin2x = 0
    2sinx - 2sinxcosx =0
    2sinx ( 1 - cos x) = 0

    2 sinx = 0 x= 0 x = 0 and 360 degrees
    1 - cos x = 0 cos x = 1 x = 0 and 360 degrees.

    Can you help I think I have missed a value between 0 and 360?

    Thanks
    Cromlix
    sin(x+2x) \neq sin(x)+sin(2x)

    try to use sin(a+b)=sin (a) cos (b)+cos (a) sin (b)
    to expand sin (x+2x)


    --hope it'll help--
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  3. #3
    Super Member bigwave's Avatar
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    honolulu
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    3\sin{\left(x\right)} = \sin{\left(3x\right)}

    3\sin{\left(x\right)} = \sin{\left(x\right)}\left[2\cos{\left(2x\right)}+1\right]

    3 = 2 \cos{\left(2x\right)} + 1

    1 = \cos{\left(2x\right)}

    \cos^{-1}{\left(1\right) = 2x

    0 = 2x

    x = 0 or x = \pi n

    takes me forever to do latex...
    Last edited by bigwave; August 25th 2010 at 02:36 PM. Reason: added step
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  4. #4
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    Lexington, MA (USA)
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    Hello, cromlix!

    We know the basic identities:

    . . \sin(A + B) \:=\:\sin A\cos B + \cos A\sin B

    . . \sin2A \:=\:2\sin A\cos A

    . . \cos2A \:=\:1 - 2\sin^2\!A \:=\:2\cos^2\!A - 1 \:=\:\cos^2\!A - \sin^2\!A



    \text{Solve for }x\!:\;\;3\sin x \:=\: \sin3x .[1]

    We need a Triple-Angle Identity . . .

    \sin3x \;=\;\sin(x+2x)

    . . . . . =\;\sin x\cos2x + \cos x\sin2x

    . . . . . =\;\sin x(1 - 2\sin^2\!x) + \cos x(2\sin x\cos x)

    . . . . . =\; \sin x - 2\sin^3\!x + 2\sin x\cos^2\!x

    . . . . . =\; \sin x - 2\sin^3\!x + 2\sin x(1 - \sin^2\!x)

    . . . . . =\;\sin x - 2\sin^3\!x + 2\sin x - 2\sin^3\!x

    \sin3x \;=\; 3\sin x - 4\sin^3\!x


    Equation [1] become: . 3\sin x \;=\;3\sin x - 4\sin^3\!x

    . . 4\sin^3\!x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:0


    Therefore: . x \:=\:\pi n\;\text{ for any integer }n

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  5. #5
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    Arbroath, Bonnie Scotland.
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    Smile Trigonometric question

    Thank you all very much for your help.

    Cromlix
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