已知sin(α-3π)=2cos(α-4π),求sin(π-α)+5cos(2π-α) / [2sin(3/2π-α)-sin(-α)]

问题描述:

已知sin(α-3π)=2cos(α-4π),求sin(π-α)+5cos(2π-α) / [2sin(3/2π-α)-sin(-α)]

即-sinα=2cosα
sinα=-2cosα
原式=(sin+5cosα)/(-3cosα+sinα)
=(-2cosα+5cosα)/(-3cosα-2cosα)
=-3/5