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化简:(1)1/2cosx-√3/2sinx(2)√3sinx+cosx(3)√2(sinx-cosx)(4)√2cosx-√6sinx
人气:360 ℃ 时间:2019-09-11 13:03:29
解答
(1)1/2cosx-√3/2sinx
=sinπ/6 cosx-cosπ/6 sinx
=sin(π/6-x)
(2)√3sinx+cosx
=2(√3/2sinx+1/2 cosx)
=2(cosπ/6sinx+sinπ/6cosx)
=2sin(x+π/6)
(3)√2(sinx-cosx)
=2(√2/2sinx-√2/2cosx)
=2(cosπ/4 sinx-sinπ/4 cosx)
=2sin(x-π/4)
(4)√2cosx-√6sinx
=2√2(1/2 cosx-√3/2 sinx)
=2√2(sinπ/6 cosx-cosπ/6 sinx)
=2√2 sin(π/6-x)
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