若函数f(x)=根号5sin(2x Θ)

1,f(x)=sin²x+√3sinxcosx+2cos²x＝1－cos²x＋√3/2sin2x+2cos²x＝cos²x＋√3/2sin2x+1=(

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

1.2sin(30+2a)+4=5sin(30+2a)=1/230+2a=3030+2a=150a=0a=60tana=√32.f(x)=3sin²x+2√3sinxcosx+5cos

我刚答过的题:(I)f(x)=3sin^2x+2√3sinxcosx+5cos^2x=sin^2x+2√3sinxcosx+3cos^2x+2(sin^2x+cos^2x)=sin^2x+2√3sin

0到正无穷

f(x)=√2sin(x+π/4)∵x∈(0,π/2)∴x+π/4∈（π/4,π/2）∴sin(x+π/4)∈(√2/2,1]∴√2sin(x+π/4)∈(1,√2]即函数值域为(1,√2]

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

f(x)=2根号3sin方x+sin2x+根号3＝根号3（2sin方x＋1）＋sin2x＝根号3（1－cos2x＋1）＋sin2x＝2根号3－根号3cos2x＋sin2x＝2sin（2x－60度）＋2

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

主要用了4个公式cos^2x=(1+cos2x)/2sin^2x=(1-cos2x)/2sin2x=2sinxcosxasinx+bsinx=根号(a^2+b^2)sin(x+arctan(b/a))

f(x)=3sin^2x+2根号3sinxcosx+5cos^2x=3sin^2x+√3sin2x+5cos^2x-4cos^x-4sin^2x+4=2sin(30+2x)+4T=∏最大值为62sin

函数f(x)=负根号3sin^2x+sinxcosx应该没^这个符号的吧?如果是没有的话f(x)=负根号3sin2x+sinxcosx=负根号3sin2x+sin2x=(1/2-3^(1/2))sin

f(25π/6)=f(π/6）3sin²x+sinxcosx=3sin²x+0.5sin2xf（x)=--根号下3sin²x+0.5sin2x然后根据函数的单调性就可求出

①f(π/3-x)=f(π/3+x),说明函数图像的一条对称轴是x=π/3,而三角函数图像的对称轴必定过它的最高点或最低点,所以f(π/3)=√5或-√5.②由①知：f(π/3)=√5或-√5,即√5

1：(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2

(1)已知函数f(x)=-(√3)sin²x+sinxcosx,求f(25π/6).f(x)=-(√3)sin²x+sinxcosx=(√3/2)(cos2x-1)+(1/2)si

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin²x=[2sinxcos(π/3)+2cosxsin(π/3)+sinx]cosx-√3sin²x=(2sinx+

∵f(a)=5∴sin²a+2√3sinacosa+5cos²a=5sin²a+2√3sinacosa=5-5cos²asin²a+2√3sinaco

(sqrt是开方)f(x)=5sqrt(3)[cos(2x)]^2+sqrt(3)[sin(2x)]^2-4sin(x)cos(x),由于[cos(2x)]^2+[sin(2x)]^2=1,且2sin

1)f=2sin2xcos(Pi/3)+根号3*cos2x-m=sin2x+根号3*cos2x-m=2[sin2xcos(Pi/3)+cos2xsin(Pi/3)]-m=2sin(2x+Pi/3)-m