Question

已知△ABC中,满足三内角sin^B+sin^2C-sinBsinC=sin^2A,则A=?

Answer 1

(1)

sin^2B+sin^2C-sinBsinC=sin^2A

先用正弦定理得

b^2+c^2-bc=a^2

再用余弦定理得

cosA=(b^2+c^2-a^2)/(2bc)=bc/(2bc)=1/2

∴A=30°

(2)

y=sinxcosxcos2x

=1/2*2sinxcosxcos2x

=1/2*sin2x*cos2x

=1/4*2sin2x*cos2x

=1/4sin4x

值域是[-1/4,1/4]