Question Number 99539 by Ar Brandon last updated on 21/Jun/20
$$\mathrm{If}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:\mathrm{Prove} \\ $$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=\mathrm{4cos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Acos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Bcos}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{C} \\ $$
Answered by Dwaipayan Shikari last updated on 22/Jun/20
![sinA+sinB+sinC=2sin(A/2)cos(A/2)+2sin((B+C)/2)cos((B−C)/2) =2cos(A/2)(sin(A/2)+cos((B−C)/2)) =4cos(A/2)(cos(((π/2)−((A+C)/2)+(B/2))/2)cos(((π/2)−((A+B)/2)+(C/2))/2)) =4cos(A/2)cos(B/2)cos(C/2)[Proved]{As you can see((π/2)−((A+C)/2))=(B/2) {And((π/2)−((A+B)/2))=(C/2) ,sin(A/2)=cos((π/2)−(A/2)) {sin((C+B)/2)=cos(A/2)](https://www.tinkutara.com/question/Q99540.png)
$${sinA}+{sinB}+{sinC}=\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{B}+{C}}{\mathrm{2}}{cos}\frac{{B}−{C}}{\mathrm{2}} \\ $$$$=\mathrm{2}{cos}\frac{{A}}{\mathrm{2}}\left({sin}\frac{{A}}{\mathrm{2}}+{cos}\frac{{B}−{C}}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\left({cos}\frac{\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}+\frac{{B}}{\mathrm{2}}}{\mathrm{2}}{cos}\frac{\frac{\pi}{\mathrm{2}}−\frac{{A}+{B}}{\mathrm{2}}+\frac{{C}}{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}{cos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}\left[{Proved}\right]\left\{{As}\:{you}\:{can}\:{see}\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{C}}{\mathrm{2}}\right)=\frac{{B}}{\mathrm{2}}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{And}\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{B}}{\mathrm{2}}\right)=\frac{{C}}{\mathrm{2}}\:,{sin}\frac{{A}}{\mathrm{2}}={cos}\left(\frac{\pi}{\mathrm{2}}−\frac{{A}}{\mathrm{2}}\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{sin}\frac{{C}+{B}}{\mathrm{2}}={cos}\frac{{A}}{\mathrm{2}}\right. \\ $$
Commented by Ar Brandon last updated on 22/Jun/20
cool ��