解
√3sinx+cosx
=2[(√3/2)*sinx+(1/2)cosx]
=2(sinxcos30°+cosxsin30°)
=2sin(x+30°)
√2(sin×-cosx)
=2[(√2/2)sinx-(√2/2)cosx]
=2(cos45°sinx-sin45°cosx)
=2sin(x-45°)2[(√3/2)*sinx+(1/2)cosx] 3个2是怎么得来的？(√3sinx+cosx)除以2得 (√3/2)*sinx+(1/2)*cosx再乘2得 2[(√3/2)*sinx+(1/2)cosx] 原式值没有改变