Question

在三角行abc中已知p=(2sina,1),q=(2sinb,cisC),且p//q,求sinA

Answer 1

(1)p=(2sinA,1),q=(2sinB,cosC),且p//q,

∴sinB/sinA=cosC,

∴sinAcosC=sinB=sin(A+C),

∴cosAsinC=0,

∴cosA=0,sinA=1.

(2)f(C)=-2cos2C/(1+tanC)+1

=-2cosC(cosC-sinC)+1

=2sinCcosC+1-2(cosC)^2

=sin2C-cos2C

=√2sin(2C-π/4),

C∈(0,π/2）,

∴u=2C-π/4的值域是(-π/4,3π/4）,

∴v=sinu的值域是(-√2/2,1],

∴f(C)=√2v的取值范围是(-1,√2].