by user580502
Last Updated August 10, 2018 11:20 AM - source

Confused among determining A & B in these trigonometric formulae.. $2 Sin A Cos B = Sin (A+B)+ Sin (A-B)$ & $2 Cos A Sin B = Sin (A+B)- Sin (A-B)$

I've got $ 2 Sin (\frac{x}{2}) cos (nx)$ = $Sin (\frac{1}{2}+n)x + Sin (\frac{1}{2}-n)x $

Which is correct from the first formula..

But in my book it is given as, $ 2 Sin (\frac{x}{2}) cos (nx)= Sin (n+\frac{1}{2})x - Sin (n-\frac{1}{2})x $

Which is also correct...

then Why such a difference? Is there any method to determine A & B in these formulas

Note you've left off "$\sin$" on the second term in your original problem statement (typo).

The two are the same formula. It's just because $\sin \theta = -\sin(-\theta)$ since $\sin$ is an odd function. In particular, since $a-b=-(b-a)$, you can write $\sin(a-b)=-\sin(-(b-a))$.

August 10, 2018 11:01 AM

Note via this graph that: $$\sin(x)\equiv-\sin(-x)$$

Hence: $$\sin[(\frac12 -n)x]\equiv-\sin[(n-\frac12)x]$$

August 10, 2018 11:10 AM

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