sin2x-4分之 的最小值

请采纳谢谢f(x)=(sin^4+cos^4+sin^2xcos^2x)/2-sin2x=((sin^2x+cos^2x)^2-sin^2xcos^2x)/2-sin2x=(1-sin^2xcos^2

f(x)=√[1^2+(√3)^2]·sin(2x+arctan√3/1)=2sin(2x+π/3),x∈[-π/4,π/4]∴易得f(x)min=f(-π/4)=-1

f'(x)=-2cos2x-acosx=-2(2(cosx)^2-1)-acosx=-4(cosx)^2-acosx+2=0cosx=(a+-sqrt(a^2+32))/-8sinx=sqrt(1-(

再问：已经悟出来了，谢谢

y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√

f(x)=[(sinx)^4+(cosx)^4+(sinx)^2(cosx)^2]/(2-sin2x)={[(sinx)^2+(cosx)^2]^2-(sinx)^2*(cosx)^2}/(2-2si

f(x)=(sin^4+cos^4+sin^2xcos^2x)/2-sin2x=((sin^2x+cos^2x)^2-sin^2xcos^2x)/2-sin2x=(1-sin^2xcos^2x)/2-

Y=（sin^4x+cos^4x+sin^2xcos^2x）/（2-sin2x）=（(sin^2x+cos^2x)^2-sin^2xcos^2x)/（2-sin2x）=(1-sinxcosx)(1+s

y＝sin2x+cos2x=根号2sin(2x+π/4)所以周期=2π/2=π；最大值=根号2最小值=-根号2取最大值时,2x＋π/4=2kπ+π/22x=2kπ+π/4x=kπ+π/8取最小值时,2

y=根号2*（二分之根号2*sin2x+二分之根号2*cos2x）=根号2*（cos45°*sin2x+sin45°*cos2x）=根号2*sin（2x+45°）因为sin（2x+45°）在正负一之间

/>y=sin2x+2√2cos(π/4+x）+3=cos(2x-π/2)+2√2cos(π/4+x）+3=1-2sin²(x-π/4)-2√2sin(x-π/4）+3=4-2[sin

令π/4+x=Az则原式=sin（2A-π/2）+2√2cosA+3=-cos2A+2√2cosA+3=-2cosA^2+1+2√2cosA+3=-2（cosA-√2/2）^2+5由于cosA∈【-1

函数f(x)=根号3乘以sin2x+2cos^2x+m(x∈R)在区间02分之兀上的最小值为3,求f(x)的常数m的值解析：∵函数f(x)=√3sin2x+2cos^2x+m=√3sin2x+cos2

首先y=sin2x+2（cosx-sinx）+3,令cosx-sinx=t属于【-√2,√2】,所以y=-t^2+2t+4,由此易得其最小值为5.

令t=|sin2x|,则0≤t≤1,f(x)＝根号下（1+|sin2x|）+｜sin2x|＾4=√（1+t)+t^4为关于t在[0,1]的增函数,故当t=0时,f(x)有最小值1,当t=1时,f(x)

f(x)=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)=1/2sinxco

f（x）＝sin2x＋2（cosx）^2－1＝sin2x＋cos2x＝√2［sin2xcos（π/4）＋cos2xsin（π/4）］＝√2sin（2x＋π/4）.∵π/4≦x≦3π/4,∴π/2≦2x

(|sinx|+|cosx|)^2=(sinx)^2+(cosx)^2+2|sinxcosx|=1+|sin(2x)|记t=|sin(2x)|f(x)=√(1+|sin(2x)|)+(sin2x)^4

cos(x+4分之派)=5分之3cosxcosπ/4-sinxsinπ/4=3/5√2/2(cosx-sinx)=3/5得：cosx-sinx=3√2/5(cosx-sinx)^2=cos^2x-2s

y=2cos²x-1+1+sin2x=sin2x+cos2x+1=√2(sin2x*√2/2+cos2x*√2/2)+1=√2(sin2xcosπ/4+cos2xsinπ/4)+1=√2si