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