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

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

已tanx=4/3,求cos(2x-π/3)cos(π/3-x)-sin(2x-π/3)sin(π/3-x)
tanx=4/3,根据sin^2x+cos^2x=1,可以求出cosx=1/±√(1+tan^2x)=±3/5.
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.

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