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Question Number 95515 by bagjamath last updated on 25/May/20

If A+B+C = π, then sin^2 A+sin^2 B+sin^2 C−2 cos A cos B cos C=

$$\mathrm{If}\:{A}+{B}+{C}\:=\:\pi,\:\mathrm{then} \\ $$$$\mathrm{sin}^{\mathrm{2}} {A}+\mathrm{sin}^{\mathrm{2}} {B}+\mathrm{sin}^{\mathrm{2}} {C}−\mathrm{2}\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}= \\ $$

Answered by som(math1967) last updated on 25/May/20

sin^2 A+sin^2 B+sin^2 C−2cosAcos Bcos C =(1/2)(2sin^2 A+2sin^2 B+2sin^2 C) −2cosAcos Bcos C =(1/2)(1−cos2A)+(1/2)(1−cos2B) 1−cos^2 C −2cosAcosBcos C =2−(1/2)2cos (A+B)cos (A−B) −cos^2 C−2cosAcosBcosC =2 +cosCcos(A−B)−cos^2 C −2cosAcosBcosC ★ =2+cosC{cos(A−B)−cosC} −2cosAcosBcosC =2+cosC{cos (A−B)+cos (A+B)} −2cosAcosBcosC =2+2cosAcosBcosC−2cosAcosBcosC =2 ans ★A+B+C=π ∴cos(A+B)=−cosC

$$\mathrm{sin}^{\mathrm{2}} \mathrm{A}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{B}+\mathrm{sin}^{\mathrm{2}} \mathrm{C}−\mathrm{2cosAcos}\:\mathrm{Bcos}\:\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin}^{\mathrm{2}} \mathrm{A}+\mathrm{2sin}\:^{\mathrm{2}} \mathrm{B}+\mathrm{2sin}^{\mathrm{2}} \mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2cosAcos}\:\mathrm{Bcos}\:\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos2A}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos2B}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{C}\:−\mathrm{2cosAcosBcos}\:\mathrm{C} \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2cos}\:\left(\mathrm{A}+\mathrm{B}\right)\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right) \\ $$$$\:\:\:\:\:\:\:−\mathrm{cos}\:^{\mathrm{2}} \mathrm{C}−\mathrm{2cosAcosBcosC} \\ $$$$=\mathrm{2}\:+\mathrm{cosCcos}\left(\mathrm{A}−\mathrm{B}\right)−\mathrm{cos}^{\mathrm{2}} \mathrm{C} \\ $$$$\:\:−\mathrm{2cosAcosBcosC}\:\bigstar \\ $$$$=\mathrm{2}+\mathrm{cosC}\left\{\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)−\mathrm{cosC}\right\} \\ $$$$\:\:−\mathrm{2cosAcosBcosC} \\ $$$$=\mathrm{2}+\mathrm{cosC}\left\{\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)+\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)\right\} \\ $$$$−\mathrm{2cosAcosBcosC} \\ $$$$=\mathrm{2}+\mathrm{2cosAcosBcosC}−\mathrm{2cosAcosBcosC} \\ $$$$=\mathrm{2}\:\mathrm{ans} \\ $$$$\bigstar\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ $$$$\therefore\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)=−\mathrm{cosC} \\ $$

Commented by peter frank last updated on 25/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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