Question

已知锐角三角形ABC中,边A,B,C分别是角,B,C所对的边,若满足(a+b+c)(b+c-a)=3bc求 cosB+c

Answer 1

原式=(b+c)^2-a^2=3bc.带入

（sinb+sinc)^2-sin(b+c)^2=3sinb*sinc

即(sinb)^+(sinc)^2-sinb*sinc

=2[(sinb*cosc)^2+(sinc*cosb)^2+2sinb*sinc*cosb*cosc]

=(sinb*cosc)^2+(sinc*cosb)^2+2sinb*sinc*cosb*cosc

(sinb)^+(sinc)^2-sinb*sinc-(sinb*cosc)^2+(sinc*cosb)^2

=(sinb*sinc)^2+(sinc*sinb)^2

=2(sinb*sinc*cosb*cosc)

(sinb*sinc)^2+(sinc*sinb)^2-2(sinb*sinc*cosb*cosc)=0

sinbsinc*sinb*sinc=sinb*sinc*cosb*cosc

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