三角函数高一,速回,谢怎么整理得到的?
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/16 09:28:39
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三角函数高一,速回,谢怎么整理得到的?
三角函数高一,速回,谢
怎么整理得到的?
三角函数高一,速回,谢怎么整理得到的?
解原式
=1/2[1+cos(2x-4π/3)]+1/2[1-cos(2x-5π/3)]+a/2*2sinx/2cosx/2
=1/2[1+cos(2x-4π/3)+1-cos(2x-5π/3)]+a/2*2sinx/2cosx/2
=1/2[2+cos(2x-4π/3)-cos(2x-5π/3)]+a/2sinx
=1+1/2[cos(2x-4π/3)-cos(2...
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解原式
=1/2[1+cos(2x-4π/3)]+1/2[1-cos(2x-5π/3)]+a/2*2sinx/2cosx/2
=1/2[1+cos(2x-4π/3)+1-cos(2x-5π/3)]+a/2*2sinx/2cosx/2
=1/2[2+cos(2x-4π/3)-cos(2x-5π/3)]+a/2sinx
=1+1/2[cos(2x-4π/3)-cos(2x-5π/3)]+a/2sinx
=1+1/2[cos2xcos4π/3+sin2xsin4π/3-cos2xcos5π/3-sin2xsin5π/3]+a/2sinx
=1+1/2[cos2xcos4π/3-cos2xcos5π/3+sin2xsin4π/3-sin2xsin5π/3]+a/2sinx
=1+1/2[-cos2xcosπ/3+cos2xcos2π/3-sin2xsinπ/3+sin2xsin2π/3]+a/2sinx
=1+1/2[-cos2x*1/2+cos2x(-1/2)-sin2xsin2π/3+sin2xsin2π/3]+a/2sinx
=1+1/2[-cos2x*1/2+cos2x(-1/2)]+a/2sinx
=1+1/2[-cos2x]+a/2sinx
=1+1/2[-(1-2sin^2x)]+a/2sinx
=1+1/2[-1+2sin^2x]+a/2sinx
=1-1/2+sin^2x+a/2sinx
=sin^2x+a/2sinx+1/2.
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