题目

已知A.B.C.是三角形A.B.C的三个内角,向量m=(-2,1),n=(cos(A+π/6),sin(A-π/3)),且m垂直n
求角A 若sin²C-cos²C/（1-sin2C）=-2,求tanB的值

提问时间：2020-08-11

答案

1
m=(-2,1),n=(cos(A+π/6),sin(A-π/3))
m⊥n,即：m·n=(-2,1)·(cos(A+π/6),sin(A-π/3))
=-2cos(A+π/6)+sin(A-π/3)
=-2(√3cosA/2-sinA/2)+(sinA/2-√3cosA/2)
=3sinA/2-3√3cosA/2
=3sin(A-π/3)=0
即：sin(A-π/3)=0
A∈(0,π),即：A-π/3∈(-π/3,2π/3)
即：A-π/3=0
即：A=π/3
2
(sin²C-cos²C)/(1-sin2C)
=(sinC+cosC)(sinC-cosC)/(sinC-cosC)^2
=(sinC+cosC)/(sinC-cosC)=-2
即：3sinC=cosC
即：tanC=1/3
B+C=2π/3,即：B=2π/3-C
即：tanB=tan(2π/3-C)
=(tan(2π/3)-tanC)/(1+tan(2π/3)tanC)
=(-√3-1/3)/(1-√3/3)
=-(6+5√3)/3