Question

数学第二大题求解答

数学第二大题求解答求解附过程

Answer 1

cos(θ-3π/2)=cos(3π/2-θ)=cos[π+(π/2-θ)]=
=-cos(π/2-θ)=-[-sinθ]=sinθ

sin(7π/2+θ)=sin[3π+(π/2+θ)]=
=-sin(π/2+θ)=-cosθ

sin(-θ-π)=-sin(π+θ)=-[-sinθ]=sinθ

所以，f(θ)=cos(θ-3π/2)sin(7π/2+θ)/sin(-θ-π)=sinθ(-cosθ)/sinθ=-cosθ

(1) f(θ)=-cosθ=1/3，则cosθ=-1/3，
tan²θ=sin²θ/cos²θ=(1-cos²θ)/cos²θ=
=(1-1/9)/(1/9)=8
tanθ=±√8=±2√2。

(2) f(5π/6+θ)=-cos(5π/6+θ)=-[-cos[π-(5π/6+θ)]]=-[-cos(π/6-θ)]=-f(π/6-θ)=-1/3