Prove $\displaystyle
(sinA+sinB)(sinA-sinB)=sin(A+B)sin(A-B)
$
Hello, Punch!
Start with the right side . . .Prove: .$\displaystyle (\sin A+\sin B)(\sin A-\sin B)\:=\:\sin(A+B)\sin(A-B)$
$\displaystyle \sin(A\;+\;B)\sin(A\;-\;B) \;=\;\bigg[\sin A\cos B\;+\;\cos A\sin B\bigg]\,\bigg[\sin A\cos B\;-\;\cos A\sin B\bigg]$
. - - . . . . . . . . . . . $\displaystyle =\; \sin^2\!A\cos^2\!B \;\;-\;\; \cos^2\!A\sin^2\!B $
. . . . . . . . . . . $\displaystyle = \; \sin^2\!A\overbrace{(1-\sin^2\!B)} - \overbrace{(1-\sin^2\!A)}\sin^2\!B$
. . . . . . . . . .$\displaystyle =\; \sin^2\!A - \sin^2\!A\sin^2\!B - \sin^2\!B + \sin^2\!A\sin^2\!B $
. . . . . . . . . . . . . . . . . . $\displaystyle = \;\sin^2\!A - \sin^2\!B$
. . . . . . . . . . . . . .$\displaystyle = \;(\sin A + \sin B)\,(\sin A - \sin B)$
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