Question

已知sina=2/3,a∈(π/2,π),求cos(π/3+a),cosa(π/3-a)

Answer 1

我觉得题目打错了吧,应该是cos(π/3-a)?

a∈(π/2,π),则cosa=-√[1-(sina)^2]=-√5/3

cos(π/3+a)=cos(π/3)cosa-sin(π/3)sina=(1/2)(-√5/3)-(√3/2)(2/3)=-(√5+2√3)/6

cos(π/3-a)=cos(π/3)cosa+sin(π/3)sina=(1/2)(-√5/3)+(√3/2)(2/3)=(2√3-√5)/6