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Math Help - proof

  1. #1
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    proof

    i need to prove sinx-siny= 2 sin( (x - y)/2 ) cos( (x + y)/2 )
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  2. #2
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    here is a simple say.

    \sin(a)+\sin(b)=\sin(x+y)+\sin(x-y) for some x and y.

    after applying the sine addition and subtraction formulas,

    \sin(x\pm y)=\sin(x)\cos(y)\pm\sin(y)\cos(x), you will end up with

    2\sin(x)\cos(y) now note that we want this in terms of a and b so we have

    x=\frac{a+b}{2} and y=\frac{a-b}{2} so using this we have

    \sin(a)+\sin(b)=2\sin\left(\frac{a+b}{2}\right)\co  s\left(\frac{a-b}{2}\right)
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  3. #3
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    Hello, lexmex1546!

    This is a rehash of putnam120's solution . . . with baby steps.


    Prove: . \sin x - \sin y \:= \:2\sin\left(\frac{x - y}{2}\right)\cos\left(\frac{x + y}{2}\right)

    We have these two formulas:

    . . (1)\;\sin(a + b) \:=\:\sin a\cos b + \sin b\cos a
    . . (2)\;\sin(a-b) \:=\:\sin a\cos b - \sin b\cos a

    Subtract (2) from (1): . \sin(a+b) - \sin(a-b) \:=\:2\sin b\cos a\;\;(3)

    Let: . \begin{Bmatrix}x\:=\:a+b \\ y\:=\:a-b\end{Bmatrix}\;\;\Rightarrow\;\;\begin{Bmatrix}a \:=\:\frac{x+y}{2} \\ b \:= \:\frac{x-y}{2}\end{Bmatrix}

    Then (3) becomes: . \boxed{\sin x - \sin y \:=\:2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)}

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