ldcf.net
当前位置:首页 >> 已知函数F(x)4sin²(π/4+x)%2√3Cos2x%1,x∈... >>

已知函数F(x)4sin²(π/4+x)%2√3Cos2x%1,x∈...

∵f(x)=4sin^(π/4+x)-2√3cos2x-1 ∴f(x)=2[1-cos(π/2+2x)]-2√3cos2x-1 =2sin2x-2√3cos2x+1 =4sin(2x-π/6)+1 又π/2≤2x≤π,π/3≤2x-π/6≤5/6π ∴f(x)max=5 f(x)min=3 若q|f(x)-m|

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

解答:(1)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22x-π/3=π/2时,f(x)有最大值3(2)|f(x)-m|m-2且f(x)m-2...

(1)由题意知,f(x)=2sin2(π4+x)?3cos2x?1=1?cos(π2+2x)?3cos2x?1=sin2x?3cos2x=2sin(2x?π3),∴h(x)=f(x+t)=2sin(2x+2t?π3),∴h(x)的图象的对称中心为(kπ2+π6?t,0),k∈Z,又∵已知点(?π6,0)为h(x)的图象的一个对称中心,∴t=kπ2+...

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...

f(x)=2sin平方(π/4+x)-根号3cos2x,x∈(π/4,π/2) =[1-cos(2x+π/2)]-√3cos2x =1+sin2x-√3cos2x =1+2sin(2x-π/3) f(X)的最大值3和最小值1 x∈(π/4,π/2),y∈(1/2,1) │f(x)-m│<2 f(x)-2

(1)f(x)=1-cos(π/2+2x)-√3cos2x =1-sin2x-√3cos2x =1-2sin(2x+π/3) f(a)=1-2sin(2a+π/3)=1/3 所以sin(2a+π/3)=1/3 所以f(α+π/6)=1-2sin(2α+π/3)=1/3. (ii)g(x)=a[1-2sin(2x+π/3)]+b 定义域为[0,π/2],2x+π/3∈[π/3,4π/3], sin(2x+π/3)∈[-√3/2,...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

(PS:2√3=2倍根号3) 解: (1) ∵sin^2(π/4+x)=[1-cos(π/2+2x)]/2 (降幂公式 : sinx的平方 = [1-cos2x]/2 ) ∴f(x)=4sin^2(π/4+x)-2√3cos2x-1=2+2sin2x-2√3cos2x-1 (公式: cos(π/2+2x)=-sin2x) ∴f(x)=1+4sin(2x-π/3) (三角函数角归一公式) ∵-1...

(1)f【4/π】=-(a+1)sinθ=0, ∵θ∈(0,π). ∴sinθ≠0, ∴a+1=0,即a=-1 ∵f(x)为奇函数, ∴f(0)=(a+2)cosθ=0, ∴cosθ=0,θ=2/π。 (2)由(1)知f(x)=(-1+2cos2x)cos(2x+2/π)=cos2x•(-sin2x)=-2/1sin4x, ∴f(4/a)=-2/1 ...

网站首页 | 网站地图
All rights reserved Powered by www.ldcf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com