化简：（1）1/2cosx-√3/2sinx（2）√3sinx+cosx（3）√2(sinx-cosx)（4）√2cosx-√6sinx
问题描述：
化简：（1）1/2cosx-√3/2sinx（2）√3sinx+cosx（3）√2(sinx-cosx)（4）√2cosx-√6sinx
答
1.1/2cosx-√3/2sinx
=cosπ/3cosx-sinπ/3sinx
=cos(x+π/3)
2.√3sinx+cosx
=2[(√3/2)sinx+(1/2)cosx]
=2(cosπ/6sinx+sinπ/6cosx)
=2sin(x-π/6)
3.√2(sinx-cosx)
=2[(√2/2)sinx-(√2/2)cosx]
=2(cosπ/4sinx-sinπ/4cosx)
=2sin(x-π/4)
4.√2cosx-√6sinx
=2√2[(1/2)cosx-(√3/2)sinx]
=2√2(cosπ/3cosx-sinπ/6sinx)
=2√2cos(x+π/3)