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化简函数f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx 0分
人气:462 ℃ 时间:2020-03-31 06:54:00
解答
f(x)=2cosx.cos(x-π/6)-√3(sinx)^2+sinxcosx
=2cosx*(cosx*cosπ/6+sinx*sinπ/6)-√3(sinx)^2+sinx*cosx
=2cosx*(√3/2*cosx+1/2*sinx)-√3(sinx)^2+sinx*cosx
=√3(cosx)^2+sinx*cosx-√3(sinx)^2+sinx*cosx
=√3((cosx)^2-(sinx)^2)+2sinx*cosx
=√3cos2x+sin2x
=2sin(2x+π/3)
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