若cos⊙=1/5,且3/2π<x<2π,那么:
sin⊙=-根号(1-cos平方⊙)=-根号(1- 1/25)=-2(根号6)/5
所以:
cos(⊙-π/3)
=cos⊙cos(π/3) +sin⊙sin(π/3)
=(1/5)*(1/2)+[-2(根号6)/5]*(根号3)/2
=(1- 6根号2)/10
若cos⊙=1/5,且3/2π<x<2π,那么:
sin⊙=-根号(1-cos平方⊙)=-根号(1- 1/25)=-2(根号6)/5
所以:
cos(⊙-π/3)
=cos⊙cos(π/3) +sin⊙sin(π/3)
=(1/5)*(1/2)+[-2(根号6)/5]*(根号3)/2
=(1- 6根号2)/10