求函数f(x)=sinx cosx在x∈{-2分之排}
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(1)化简可得f(x)=4sinx(cosxcosπ3-sinxsinπ3)+3=2sinxcosx-23sin2x+3=sin2x+3cos2x…(2分)=2sin(2x+π3)…(4分)所以T=2
f(x)=sinxcosΦ+cosxsinΦ=sin(x+Φ)因为函数y=f(2x+π/4)的图像关于x=π/6对称所以y=f(2x+π/4)=sin(2x+π/4+Φ)的图像关于x=π/6对称所以s
请问,会不会多了一个cosx?
cos²x=1-sin²x=(1+sinx)(1-sinx).∴y=2sinx(1-sinx)=-2(sin²x-sinx)=-2[sinx-(1/2)]²+(
原式=2sinxcos(x+π/3)+√3cos²x+sinxcosx=2sinxcos(x+π/3)+cosx(√3cosx+sinx)=2sinxcos(x+π/3)+2cosx·sin
f(x)=2(cosx)^2-2√3sinxcosx-1=(cos2x+1)-√3sin2x-1=cos2x-√3sin2x=2cos(2x+π/6)周期T=2π/│ω│=2π/2=π因为y=cosx
f(x)=(1+cos2x)/2+(√3/2)sin2x+3/2=(√3/2)sin2x+(1/2)cos2x+2=sin2xcos(π/6)+cos2xsin(π/6)+2=sin(2x+π/6)+
(1)最简单的方法是用“积化和差”公式2sinαcosβ=sin(α+β)+sin(α-β)原式=2×2sinxcos(x+π/3)=2[sin(x+x+π/3)+sin(x-x-π/3)]=2[si
f(x)=2sinxcos^2φ/2+cosxsinφ-sinx=f(x)=sinx(2cos^2φ/2-1)+cosxsinφ=sinxcosφ+cosxsinφ=sin(x+φ)在x=π处取得最小
y'=cosx-3sin²xcosx
y=sin方x+sinxcos(派/6-x)=(3/2)sin²x+(√3/2)sinxcosx=(√3/2)sin(2x-π/3)+3/4周期为π增区间为[kπ-π/12,kπ+5π/12
1.f(x)=sinxcosφ+cosxsinφ=sin(x+φ),周期为2∏2.y=f(2x+∏/4)=sin(2x+∏/2+φ)=-cos(2x+φ)=-cos(2(x+φ/2))cost的图像关
f(x)=sinxcosφ+cosxsinφ=sin(x+φ),sin(∏/6+φ)=1,φ=∏/3+2k∏;y不变,x缩小半;
1、f(x)=sinxcos幻+cosxsin幻=sin(x+幻)所以T=2π/1=2π2、把点代入y=f(2x+π/6)f(2x+π/6)=sin(2x+π/6+幻)1/2=sin(π/2+幻)=c
正在解答再问:答案呢我问你再答:放心,包正确再答:正在解答啊再问:答案给我就给好评再问:要全面再答:再答:再做第二问再答:合作愉快再答:把横坐标变为原来的二分之一再问:亮一点再答:图像向左平移六分之派
f(x)=2sinxcos^θ/2+cosxsinθ-sinx=sinx(cosθ+1)+cosxsinθ-sinx=sinxcosθ+cosxsinθ+sinx-sinx=sin(x+θ)f(A)=
y=2sinxcos^2x/1-sinx=2sinx(1+sinx)=2(sinx+1/2)^2-1/2-1/2≤sinx+1/2≤3/20≤(sinx+1/2)^2≤9/4函数y=2sinxcos^
解原式=2sinxcos(x+π/3)+根号3cos的平方x+1/2sin2x=2sinxcos(x+π/3)+根号3cos的平方x+sinxcosx=2sinxcos(x+π/3)+cosx(根号3
f(x)=2cosxsin(x+π/6)+2sinxcos(x+π/6)=2sin(2x+π/6),(1)x∈[0,π/6],∴2x+π/6∈[π/6,π/2],∴f(x)的值域是[1,2].(2)f
由题有:f(x)=2sin(2x+兀/6)因为:x属于[0,丌/2]所以:2x+兀/6属于[丌/6,7丌/6]所以:f(x)值域为:[-1,2]