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Math Help - sry for the trouble! another trigo problem!!

  1. #1
    cyy
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    Exclamation sry for the trouble! another trigo problem!!

    Hi sry for the previous double post, below is another porblem i faced:
    given: 0 <= x < pi ; 0 <= y; < pi
    ...


    sin^2x = sin^2y
    (sinx + siny )( sinx - siny ) = 0
    thus, how to get from above equation to form below 2 statements?
    x = y+2pi( k ) //where is the pi & k come from?
    x = -y+2pi( k )



    Thank You!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cyy View Post
    Hi sry for the previous double post, below is another porblem i faced:
    given: 0 <= x < pi ; 0 <= y; < pi
    ...


    sin^2x = sin^2y
    (sinx + siny )( sinx - siny ) = 0
    thus, how to get from above equation to form below 2 statements?
    x = y+2pi( k ) //where is the pi & k come from?
    x = -y+2pi( k )



    Thank You!
    There is a problem with one of the solutions:

    Given two real numbers a and b, if ab = 0 then a = 0 and/or b = 0.

    So starting with
    (sin(x) + sin(y) )( sin(x) - sin(y) ) = 0

    We see that
    sin(x) + sin(y) = 0
    or
    sin(x) - sin(y) = 0

    The second is (almost) trivial, so I'll do the first.
    sin(x) + sin(y) = 0

    sin(x) = -sin(y)

    This can only happen when x = y + (2k + 1) \pi. If you don't see why then expand it:
    sin(y + (2k + 1) \pi) = sin(x)~cos((2k + 1) \pi) + sin((2k + 1) \pi)~cos(y)

    Now, 2k + 1 is odd, so cos((2k +1) \pi) = -1 and sin((2k + 1)\pi) = 0.

    Thus
    x = y + (2k + 1) \pi

    The other solution
    x = y + 2k \pi
    is correct.

    -Dan
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