若sin(π+θ)=-3/5,θ是第二象限角,sin(π/2+φ)=-2根号5/5,φ是第三象限角,求cos(θ-φ)

∵sin(π+θ)=-3/5,∴-sinθ=-3/5,sinθ=3/5又θ是第二象限角∴cosθ=-√(1-sin²θ)=-4/5∵sin(π/2+φ)=-2根号5/5,∴cosφ=-2√5/5∵φ是第三象限角,∴sinφ=-√(1-cos²φ)=-√5/5∴cos(θ-φ) =cosθcosφ+sinθsinφ=-4/5*(-2√5/5)+3/5*(-√5/5)=√5/5