问题描述:
cos(x-6分之π)=-3分之根3,则cosx+cos(x-3分之π)=
最佳答案:
cosθ+cosφ = 2 cos[(θ+φ)/2] cos[(θ-φ)/2]
cosx+cos(x-π/3)
=2 cos[(x+x-π/3)/2] cos[(x-x+π/3)/2]
=2 cos(x-π/6) cosπ/6
=2(-3分之根3)(2分之根3)
=-1
cos(x-6分之π)=-3分之根3,则cosx+cos(x-3分之π)=
问题描述:
cos(x-6分之π)=-3分之根3,则cosx+cos(x-3分之π)=
cosθ+cosφ = 2 cos[(θ+φ)/2] cos[(θ-φ)/2]
cosx+cos(x-π/3)
=2 cos[(x+x-π/3)/2] cos[(x-x+π/3)/2]
=2 cos(x-π/6) cosπ/6
=2(-3分之根3)(2分之根3)
=-1