Question

已知A、B、C为△ABC的三个内角,=(sinB+cosB,cosC), =(sinC,sinB-cosB).

(1)若·=0,求角A;

(2)若·=-,求tan2A.

Answer 1

解：(1)由已知·=0,得(sinB+cosB)sinC+cosC(sinB-cosB)=0,

化简得sin(B+C)-cos(B+C)=0,即sinA+cosA=0.∴tanA=-1.而A∈(0,π),∴A=.

(2) ·=-sin(B+C)-cos(B+C)=-.∴sinA+cosA=-.

平方得2sinAcosA=-,

∵-＜0,∴A∈(,π).sinA-cosA==.

联立得sinA=,cosA=-,

∴tanA=-.∴tan2A==-.