求函数f(x)=cos^2(x-pai/12)+sin^2(x+pai/12)-1的周期
求函数f(x)=cos^2(x-pai/12)+sin^2(x+pai/12)-1的周期
教我一下过程,高一的基本忘了,公式最好也列出来!
谢谢了,各位大虾!
=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1
=[cos(2x-π/6)-cos(2x+π/6)]/2
=(sin2x)/2
所以最小正周期为π
这里主要公式是二倍角公式和两角和差公式，都是常规公式，书上有
参考资料：团队：我最爱数学！
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sin²x=(1-cos2x)/2
所以f(x)=[1+cos(2x-π/6)]/2+[1-cos(2x+π/6)]/2-1
=1/2[cos(2x-π/6)-cos(2x+π/6)]
=1/2[(cos2xcosπ/6+sin2xsinπ/6)-(cos2xcosπ/6-sin2xsinπ/6)]
=sin2xsinπ/6
=(1/2)sin2x
所以T=2π/2=π本回答被提问者采纳
求函数f(x)=cos^2(x-pai\/12)+sin^2(x+pai\/12)-1的周期
f(x)=cos^2(x-pai\/12)+sin^2(x+pai\/12)-1 =[1+cos(2x-π\/6)]\/2+[1-cos(2x+π\/6)]\/2-1 =[cos(2x-π\/6)-cos(2x+π\/6)]\/2 =(sin2x)\/2 所以最小正周期为π 这里主要公式是二倍角公式和两角和差公式，都是常规公式，书上有 参考资料：团队：我最爱数学！
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