Question

在△ABC中，化简sin^2B+sin^2C-2cosAsinBsinC

Answer 1

sin^2B+sin^2C-2cosAsinBsinC
=sin²B+sin²C+2cos(B+C)sinBsinC
=sin²B+sin²C+2（cosBcosC-sinCsinB）sinBsinC
=sin²B+sin²C+2cosBcosCsinBsinC-2sin²Bsin²C
=sin²B(1-sin²C)+sin²C(1-sin²B)+2cosBcosCsinBsinC
=sin²Bcos²C+sin²Ccos²B+2cosBcosCsinBsinC
=（sinBcosC+cosBsinC）²
=sin²（B+C）
=sin²A

追问

要化简，化简呀

Answer 2

原式=sin^2B+sin^C+2cos(B+C)sinBsinC
=sin^2B+sin^C+2(cosBcosC-sinBsinC)sinBsinC
=sin^2B+sin^C+2cosBcosCsinBsinC-2sin^2Bsin^2C
=sin^2B-sin^2Bsin^2C+sin^2C-sin^2Bsin^2C+2cosBcosCsinBsinC
=sin^2Bcos^2C+sin^2Ccos^2B+2cosBcosCsinBsinC
=(sinBcosC+sinCcosB)^2
=sin^2(B+C)
=sin^2A

Answer 3

(sinA)^2+(sinB)^2+(sinC)^2=2+2cosAcosBcosC
(sinB)^2+(sinC)^2=2+2cosAcosBcosC-(sinA)^2

sin^2B+sin^2C-2cosAsinBsinC
=2+2cosAcosBcosC-(sinA)^2-2cosAsinBsinC
=2+2cosAcosBcosC-2cosAsinBsinC-(sinA)^2
=2+2cosA(cosBcosC-sinBsinC)-(sinA)^2
=2+2cosAcos(B+C)-(sinA)^2
=2+2cosAcos(180-A)-(sinA)^2
=2-2(cosA)^2-(sinA)^2
=2(sinA)^2-(sinA)^2
=(sinA)^2