已tanx=4/3,求cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x)

问题描述:

已tanx=4/3,求cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x)

因为tanx=4/3所以cosc=3/5(可根据画直角三角形得出)
原式=cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x) =cos[(2x-π/3)+(π/3-x)]=cosx=3/5