Question

一道数学题：数列+三角函数在三角形ABC中,若tanB=cos(B-C)/[sinA-cos(B-C)],求证：(b+c

Answer 1

sinA+sin[(C-B)]=2sin[(A+C-B)/2]cos[A+B-C)/2]

=2sin[(180°-2B)/2]cos[(180°-2C)/2]

=2sin(90°-B)cos(90°-C)

=2cosBsinC

∴tanB=[cos(C-B)]/[sinA+sin(C-B)]=[cos（C-B）]/2cosBsinC

2sinBsinC=cosBcosC+sinBsinC

cosBcosC-sinBsinC=0

cos（B+C）=0

B+C=90°

(b+c)/a=(sinb+sinc)/sina=sinb+sinc

=sinb+sin(90-b)

=√2(sin(b+45))

小于=√2

大于1不用证了吧!