求证[tan(2π-α)sin(-2π-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
问题描述:
求证[tan(2π-α)sin(-2π-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
答
这都是诱导公式tan(2π-α)=tan(-α)=-tanαsin(-2π-α)=sin(-α)=-sinαcos(6π-α)=cos(-α)=cosαsin(α+3π/2)=-cosαcos(α+3π/2)=sinα所以原式=(-tanα)(-sinα)cosα/[(-cosα)sinα]=-tanα