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函数y=sin2(x+π/4)%Cos2(x+π/4)

因为sin(π/2+x)=cos x,这是换角公式 所以sin(x+π/4)=sin(π/2+(x-π/4))=cos(x-π/4) 又因为cos(-x)=cos x 所以cos(π/4-x)=cos(-(x-π/4))=cos(x-π/4) 所以sin(x+π/4)=cos(π/4-x) 所以他们的平方也相等啦

使用降幂公式(sinx)^2=(1-cos2x)/2 原式可化为y=(1-cos(2x+π/2))/2-(1-cos(2x-π/2))/2 =-cos(2x+π/2)/2+cos(2x-π/2)/2 =sin2x/2+sin2x/2 =sin2x ∴原函数是周期为π的奇函数.

sin(π/2-x)=sin2(π/4-x/2)=2sin(π/4-x/2)cos(π/4-x/2) , sin(π/2-x)=cosx , sin(π/2-x)≠ cos(π/4-x/2)

y=-cos(2x+π2)=sin2x,∵ω=2,∴T=π,∵sin(-2x)=-sin2x,则函数y为周期为π的奇函数.故选A

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

y=sin(x-pi/4)sin(pi/2-(x+pi/4)) =sin(x-pi/4)sin(pi/4-x) =-(sin(x-pi/4))^2 =(cos2x-1)/2 =cos2x/2-1/2 1)反余弦值域是[0,pi],x属于【-π/4,π/4】 2x...[-pi/2,pi/2] 2x=arccos(1+2y)+pi/2......[0,pi] x=1/2arccos(1+2y)+pi/4 2) x属于【π/4...

y=2cos(x+π/4)*cos(x-π/4)+(√3)sin2x =2sin(π/4-x)*cos(π/4-x)+(√3)sin2x =sin(π/2-2x)+(√3)sin2x =cos2x+(√3)sin2x =2[cosπ/3cos2x+sinπ/3sin2x] =2cos(2x-π/3) 所以,值域[-2,2],周期为π, 2kπ

f(x)=2sin²(π/4+x)-根号3cos2x =1-cos(π/2+2x)-√3cos2x =sin2x-√3cos2x+1 =2sin(2x-π/3)+1 ∵ x∈[π/4,π/2] ∴ 2x-π/3∈[π/6,2π/3] ∴ sin(2x-π/3)∈[1/2,1] ∴ 2x-π/3=π/6时,f(x)有最小值2 2x-π/3=π/2时,f(x)有最大值3

解: (1) tanx有意义,x≠kπ+ π/2,(k∈Z) 函数定义域为{x|x≠kπ+ π/2,k∈Z} f(x)=4tanxsin(π/2 -x)cos(x- π/3) -√3 =4tanxcosxcos(x-π/3)-√3 =4sinx[cosxcos(π/3)+sinxsin(π/3)] -√3 =4sinx[(1/2)cosx+(√3/2)sinx] -√3 =2sinxcosx+2√3sin²x-√...

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