题目
已知a b c为三角形abc的三内角,向量m=(-2,-1)向量n=(cos(A+π/6),sin(A-π/3))且向量m⊥向量n
(1)求角A (2)若 (sin²C-cos²C)/1-sin2c=-2 求tanb的值
(1)求角A (2)若 (sin²C-cos²C)/1-sin2c=-2 求tanb的值
提问时间:2021-02-12
答案
(1)由向量m⊥向量n可得:
-2cos(A+π/6)-sin(A-π/3)=0,
∵cos(A+π/6)=cos[π/2+(A-π/3)]=-sin(A-π/3)
∴sin(A-π/3)=0,
∵A∈(0,π),∴A-π/3=0,即A=π/3.
(2)(sin²C-cos²C)/(1-sin2C)=-2
即(sin²C-cos²C)/(sinc-cosc)²=-2,
∴(sinC+cosC)/(sinC-cosC)=-2,
∴(tanC+1)/(tanC-1)=-2,
解得:tanC=1/3.
tanB=tan[π-(A+C)]=-tan(A+C)=-(tanA+tanc)/(1-tanAtanC)=-(6+5√3)/3
-2cos(A+π/6)-sin(A-π/3)=0,
∵cos(A+π/6)=cos[π/2+(A-π/3)]=-sin(A-π/3)
∴sin(A-π/3)=0,
∵A∈(0,π),∴A-π/3=0,即A=π/3.
(2)(sin²C-cos²C)/(1-sin2C)=-2
即(sin²C-cos²C)/(sinc-cosc)²=-2,
∴(sinC+cosC)/(sinC-cosC)=-2,
∴(tanC+1)/(tanC-1)=-2,
解得:tanC=1/3.
tanB=tan[π-(A+C)]=-tan(A+C)=-(tanA+tanc)/(1-tanAtanC)=-(6+5√3)/3
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