Question

急！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)

Answer 1

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

Answer 2

刚开始还以为是集合，原来是向量~
a
b
=sinbsinc+cosbsinc+sinbcosc-cosbcosc=(sinbcosc+cosbsinc)-(cosbcosc-sinbsinc)=sin(b+c)-cos(b+c)=sin(π-a)-cos(π-a)=sina+cosa=0
所以tana=-1(显然cosa≠0)
又a∈(0,π)，告判没故a=3π/4
a
b
=sina+cosa=-1/5
所以(sina+cosa)²=sin²a+cos²袜纳a+2sinacosa=1+sin2a=1/25
故sin2a=-24/冲胡25
并由a∈(0,π)得a=