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    Hey all, i have this question and i cant exactly prove it so i was hoping some1 could help me out.

    sin(a+b) + sin(a-b)=2sin(a)sin(b)
    I have to prove that but i only get as far as
    2sin(a)cos(b)
    which seems abit wrong to me.

    Thanks!
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    \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B

    You shouldn't have any cos' because they cancel out
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    It doesn't.

    Counterexample: A = pi/2, B = 0.

    Then sin(pi/2 + 0) + sin(pi/2 - 0) = 2 but not 2sin(pi/2)sin(0) = 0.

    I believe you should consider cos(A + B) and cos(A - B).
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    Hmm. That is wrong. \sin(a + b) + \sin(a - b) = 2 \sin (a) \cos (b).
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    thanks alot
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    (Original post by Swayum)
    It doesn't.

    Counterexample: A = pi/2, B = 0.

    Then sin(pi/2 + 0) + sin(pi/2 - 0) = 2 but not 2sin(pi/2)sin(0) = 0.

    I believe you should consider cos(A + B) and cos(A - B).
    I meant cosasinb cancels the other one
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    (Original post by boy)
    I meant cosasinb cancels the other one
    I wasn't replying specifically to you, but rather the OP. By "it doesn't", I meant sin(A + B) + sin(A - B) doesn't equal 2sin(A)cos(B) for all values.
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    (Original post by Swayum)
    I wasn't replying specifically to you, but rather the OP. By "it doesn't", I meant sin(A + B) + sin(A - B) doesn't equal 2sin(A)cos(B) for all values.
    Oh
 
 
 
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