Expression : 2cos[(C+D)/2]sin[(C-D)/2] Using the formula : $$sin x . cos y = frac{1}{2} [sin (x + y) + sin (x - y)]$$ --------------(i) Substituting $$(x + y) = C$$ and $$(x - y) = D$$ => $$x = frac{C + D}{2}$$ and $$y = frac{C - D}{2}$$ in equation (i), => $$sin (frac{C + D}{2}) . cos (frac{C - D}{2}) = frac{1}{2} [sin C + sin D]$$ => $$2 . cos (frac{C - D}{2}) . sin (frac{C + D}{2}) = sin C + sin D$$ Replacing D by -D, we get : => $$2 . cos(frac{C - (-D)}{2}) . sin(frac{C + (-D)}{2}) = sin C + sin(-D)$$ => $$2 . cos(frac{C + D}{2}) . sin(frac{C - D}{2}) = sin C - sin D$$
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