Can someone help me derive sum to product

**viciouspoultry** · March 31st 2008, 05:31 PM

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

**Soroban** · March 31st 2008, 08:15 PM

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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