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Math Help - Can someone help me derive sum to product

  1. #1
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    Can someone help me derive sum to product

    All I only need to see one of the formulas derived step by step. I understand how to derive product-to-sum but have no clue how to do sum-to-product.

    Thanks in advance
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  2. #2
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    Hello, viciouspoultry!

    I don't remember the derivation, so I came up with one.
    It's rather clumsy, but it's valid.


    All I only need to see one of the formulas derived step by step.

    Consider: . \sin\left(\frac{A+B}{2}\right) \;=\;\sin\left(\frac{A}{2} + \frac{B}{2}\right) \;=\;\sin\frac{A}{2}\cos\frac{B}{2} + \cos\frac{A}{2}\sin\frac{B}{2} .[1]

    Consider: . \cos\left(\frac{A-B}{2}\right) \;=\;\cos\left(\frac{A}{2} + \frac{B}{2}\right) \;=\;\cos\frac{A}{2}\cos\frac{B}{2} + \sin\frac{A}{2}\cos\frac{B}{2} .[2]


    Multiply [1] and [2]: . \sin\left(\frac{A+B}{2}\right)\cdot\cos\left(\frac  {A-B}{2}\right)

    . . = \;\sin\frac{A}{2}\cos\frac{A}{2}\cos^2\!\frac{A}{2  } \;+\; \sin^2\!\frac{A}{2}\sin\frac{B}{2}\cos\frac{B}{2} \;+\; \cos^2\!\frac{A}{2}\sin\frac{B}{2}\cos\frac{B}{2} \;+\; \sin\frac{A}{2}\cos\frac{A}{2}\sin^2\!\frac{B}{2}

    . . =\;\sin\frac{A}{2}\cos\frac{A}{2}\cos^2\!\frac{B}{  2} \;+\; \sin\frac{A}{2}\cos\frac{A}{2}\sin^2\!\frac{B}{2} \;+\; \sin^2\!\frac{A}{2}\sin\frac{B}{2}\cos\frac{B}{2} +\; \cos^2\frac{A}{2}\sin\frac{B}{2}\cos\frac{B}{2}

    . . = \;\sin\frac{A}{2}\cos\frac{A}{2}\underbrace{\left(  \sin^2\!\frac{B}{2} + \cos^2\!\frac{B}{2}\right)}_{\text{This is 1}} \;+ \;\sin\frac{B}{2}\cos\frac{B}{2}\underbrace{\left(  \sin^2\!\frac{A}{2} + \cos^2\!\frac{A}{2}\right)}_{\text{This is 1}}


    We have: . \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \;=\;\sin\frac{A}{2}\cos\frac{A}{2} + \sin\frac{B}{2}\cos\frac{B}{2}

    . . . . . . . . \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \;=\;\frac{1}{2}\sin A  + \frac{1}{2}\sin B


    Multiply by 2: . \boxed{\sin A + \sin B \;=\;2\sin\left(\frac{A+B}{2}\right)\cos\left(\fra  c{A-B}{2}\right)}

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