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sin3x-cos3x-sinx+cosx怎么解?
人气:175 ℃ 时间:2019-09-20 12:07:22
解答
sin3x-cos3x-sinx+cosx
=3sina-4sin^3a-(4cos^3a-3cosa)-sinx+cosx
=2sina-4sin^3a-4cos^3a+4cosx
=2sin(1-2sin^2x)+4cos(1-cos^2x)
=2sincos2x+4cossin^2x
=2sin(cos2x+sin2x)
(复习三角公式)
上面用到的公式:
sin(3x)=sin(x+2x)=sin2xcosx+cos2xsinx=2sinx(1-sin²x)+(1-2sin²x)sinx=3sinx-4sin^3x
cos3x=cos(2x+x)=cos2xcosx-sin2xsinx=(2cos²x-1)cosx-2(1-cos^x)cosx=4cos^3x-3cosx
sin2x=2sinx·cosx
cos=cos^2(x)-sin^2(x) =2cos^2(x)-1 =1-2sin^2(x)sin3x-cos3x-sinx+cosx=2sinx怎么解?解 sin3x-cos3x-sinx+cosx=2sinx 3sinx-4sin^3x-(4cos^3x-3cosx)-sinx+cosx=2sinx-4sin^3x-4cos^3x+4cosx=0sin^3x+cos^3x-cosx=0sin^3x+cosx(cos^2x-1)=0sin^3x+cosx(-sin^2x)=0sin^2x(sinx-cosx)=0解得x=π/4+nπ 或x=nπ (n=0.1.2.....)解毕。
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