已知A B C为三角形ABC的三个内角,向量a=(sinB+cosB,cosC) 向量b=(sinC,sinB-cosB)

a·b=(sinB+cosB)(sinC)+(cosC)(sinB-cosB)
=sinBsinB+cosBsinC+cosCsinB-cosCcosB
=-cosCcosB+sinBsinB+cosBsinC+cosCsinB
=-cos(B+C)+sin(B+C)=-1/5
又[sin(B+C)]^2+[cos(B+c)]^2=1
解得sin(B+C)=3/5 cos(B+C)=4/5(这里利用sin(B+C)>0舍去了一组解,因为B,C为三角形的内角)
sinA=sin(180-A)=sin(B+C)=3/5
cosA=-cos(180-A)=-cos(B+C)=-4/5
tanA=sinA/cosA=-3/4
tan2A=2tanA/[1-(tanA)^2]=2*(-3/4)/(1-9/16)=-24/7
cos2A=2(cosA)^2-1=2(-4/5)^2-1=7/25