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Question Number 21268 by oyshi last updated on 18/Sep/17
if A+B+C=π so proof sin A+sin B+sin C=4cos (A/2)cos (B/2)cos (C/2)
$${if}\:{A}+{B}+{C}=\pi \\ $$$${so}\:{proof}\: \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}=\mathrm{4cos}\:\frac{{A}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$
Answered by myintkhaing last updated on 18/Sep/17
L.H.S= sin A+sin B+sin C =2sin ((A+B)/2) cos((A−B)/2)+2sin (C/2) cos (C/2) =2sin((π/2)−(C/2))cos(((A−B)/2))+2sin((π/2)−((A+B)/2))cos (C/2) =2cos (C/2) cos(((A−B)/2))+2cos(((A+B)/2))cos (C/2) =2(cos ((A+B)/2)+cos((A−B)/2)) cos (C/2) =2(2cos (A/2)cos (B/2))cos (C/2) =4cos(A/2) cos (B/2) cos (C/2) # use A+B = π−C⇒((A+B)/2) = (π/2)−(C/2) sin x+ sin y = 2sin ((x+y)/2)cos ((x−y)/2) cos x+ cos y = 2cos ((x+y)/2) cos ((x−y)/2)
$$\mathrm{L}.\mathrm{H}.\mathrm{S}=\:\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C} \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\:\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}+\mathrm{2sin}\:\frac{\mathrm{C}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$=\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{C}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)+\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$=\mathrm{2cos}\:\frac{\mathrm{C}}{\mathrm{2}}\:\mathrm{cos}\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)+\mathrm{2cos}\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}+\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}\right)\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\mathrm{2cos}\:\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{B}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$=\mathrm{4cos}\frac{\mathrm{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}}\:# \\ $$$$\mathrm{use} \\ $$$$\mathrm{A}+\mathrm{B}\:=\:\pi−\mathrm{C}\Rightarrow\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}}−\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{x}+\:\mathrm{sin}\:\mathrm{y}\:=\:\mathrm{2sin}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\mathrm{cos}\:\frac{\mathrm{x}−\mathrm{y}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{y}\:=\:\mathrm{2cos}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{x}−\mathrm{y}}{\mathrm{2}} \\ $$

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