Question

在△ABC中,已知C=3π/4,cos2B=1/2+sin²A,求tanB 若BC=2,求△ABC的面积

Answer 1

cos2B=1/2+sin²A
cos2B=1-1/2cos2A
2cos2B+cos2A=2
2cos2B+cos2(π/4-B)=2
2cos2B+cos(π/2-2B)=2
2cos2B+sin2B=2
2-4sin^2B+2sinBcosB=2
sinB(cosB-2sinB)=0
cosB-2sinB=0
tanB =1/2
2)
tanB =sinB/cosB=1/2
cosB=2sinB
sin^2B+cos^2B=1
sin^2B+4sin^2B=1
sinB=根号5/5
cosB=2sinB=2根号5/5
sinA=sin(π/4-B)=sinπ/4cosB+cosπ/4sinB=根号2/2(cosB+sinB)=根号2/2*3根号5/5
=3根号10/10
a/sinA=b/sinB,b=sinBa/sinA=2根号5/5/3根号10/10=2根号2/3
S=1/2absibC=1/2*2*2根号2/3*根号2/2=2/3
S=2/3