函数fx=2cos(x-π 4)-1求最小正周期
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 01:54:57
f(x_=(cosx+sinx)(cosx-sinx)=cos²x-sin²x=cos2x所以T=2π/2=πf(α/2)=cosα=1/3sin²α+cos²
f(x)=v3sin(π-2x)-2cos^2x+1=v3sin2x-cos2x=2sin(2x-π/6),(1)、f(π/2)=2sin(5π/6)=2*(1/2)=1;(2)、最小正周期T=2π/
f(x)=cosx-cos(x+π/2)=cosx+sinx=3/4sin^2x+cos^2x+2sinxcosx=9/162sinxcosx=sin2x=9/16-1=-7/16
若cosα=3/5.α属于(3π/2,2π),sinα=-4/5把f(2α+π/3)代入fx=√2cos(x-π/12),化简原式=cos2α-sin2αcos2α-sin2α怎么化简的就不用我说了吧
f(x)=cos²x-2cos²x/2=cos²x-2*1/2*(cosx+1)=cos²x-cosx-1=(cosx-1/2)²-5/4这是复合函数
=(1/2)sin2x-(根号3/2)cos2x+(根号3/2)cos2x+(1/2)sin2x+1+cos2x-1=sin2x+cos2x=根号2sin(2x+pi/4)最小正周期为pi-pi/4再
f(x)=-sin2x-cos2x+3sin2x-cos2x=2sin2x-2cos2x=2根号2sin(2x-π/4)T=2π/2=π-π/2+2kπ≤2x-π/4≤π/2+2kπk属于Z-π/8+
fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2(α)-1fx=cos2x+根号3s
f(x)=cos(2x-4π/3)+2cos^2x=cos(2x-4π/3)+cos2x+1=2cos(2x-2π/3)cos2π/3+1=1-√3cos(2x-2π/3)1.当cos(2x-2π/3
再答:这是高一的题目吧再答:不谢,复习加油
原式=1/2COSX+asin(x/2)cos(x/2)=1/2COSX+a/2sinx=1/2(cosx+asinx)因为最大值是2所以(√1+a^2)/2=2a=+-√15
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin(x-π/
f(x)=√3cos(π/2-2x)+2cos^2x+2=√3sin2x+(1+cos2x)+2=√3sin2x+cos2x+3=2(√3/2sin2x+1/2cos2x)+3=2sin(2x+π/6
设函数fx=2cos^2(π/4-x)+sin(2x+π/3)-1=cos(PI/2-2x)+sin(2x+PI/3)=sin(2x)+sin(2x)/2+cos(2x)*sqrt(3)/2=sqrt
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=(1/2)cos2x+(√3/2)sin2x+(cos(π/2)-cos2x)=-(1/2)cos2x+(√3/2)sin
(1)∵cos2x=2cos^2x-1∴f(x)=1/2+cos(2x+π/6)/2对称轴2x0+π/6=π+2kπx0=5π/12+kπg(x0)=1+1/2sin(5π/6+2kπ)=5/4(2)
①f(x)=cos﹙2x-4π/3﹚+2cos²x=cos2xcos4π/3+sin2xsin4π/3+1+cos2x=1/2cos2x-√3/2sin2x+1=cos(2x+π/3)+1当
f(x)=cos²2x-sin²2x+sin4x=cos4x+sin4x=√2[(√2/2)cos4x+(√2/2)sin4x]=√2sin(4x+π/4)所以,最小正周期T=2π