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Math Help - sin(u)cos(v) != cos(v)sin(u)??

  1. #1
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    sin(u)cos(v) != cos(v)sin(u)??

    I need help understanding these trig identities:

    sin(a)cos(b) = (sin(a+b) + sin(a-b))/2

    Where I see this identity I also see it followed up by this identity:

    cos(b)sin(a) = (sin(b+a) - sin(b-a))/2

    Huh? I recall that:

    sin(-u) = -sin(u)

    So cos(b)sin(a) becomes:

    cos(b)sin(a) = (sin(b+a) + sin(a-b))/2

    Right? but this is NOT equal to the sin(a)cos(b) equation above. Shouldn't it be?

    I know that xy=yx so shouldn't cos(b)sin(a) = sin(a)cos(b)?

    Where is my logic flawed...
    Last edited by sadmath; October 11th 2010 at 11:38 AM. Reason: cleaned up a bit
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sadmath View Post
    I need help understanding these trig identities:

    sin(a)cos(b) = (sin(a+b) + sin(a-b))/2

    Where I see this identity I also see it followed up by this identity:

    cos(b)sin(a) = (sin(b+a) - sin(b-a))/2

    Huh? I recall that:

    sin(-u) = -sin(u)

    So cos(b)sin(a) becomes:

    cos(b)sin(a) = (sin(b+a) + sin(a-b))/2

    Right? but this is NOT equal to the sin(a)cos(b) equation above. Shouldn't it be?

    I know that xy=yx so shouldn't cos(b)sin(a) = sin(a)cos(b)?

    Where is my logic flawed...
    i don't see your problem. if you make the change you spoke about in the second equation, you get the first. if you make it in the first, you get the second. they are equal.
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  3. #3
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    I just went through the equality. I apologize for wasting your time, and I am a little embarrassed.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sadmath View Post
    I just went through the equality. I apologize for wasting your time, and I am a little embarrassed.
    haha, don't worry about it. our little secret
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