若SIN 2X π 6=-1 3,则SIN17π 6 2X

2sin2x+cos2x=1,所以4sinx*cosx=2(sinx)^2得出x=0或则tanx=2,当x=0时,[2(cosx)^2+sin2x]/(1+tanx)=2当tanx=2时,[2(cos

sin^2x+cos^2x=1sin^2x+9sin^2x=1sin^2x=1/10Cos2X+Sin2X=cos^2x-sin^2x+2sinxcosx=9sin^2x-sin^2x+6sin^2x

f(x+π/6)=a×b=cos2xsinx+sin2xcosx=sin3xf(x)=sin[3(x-π/6)]=sin(3x-π/2)因此f(x)的最小正周期＝2π/3函数取最大值时3x-π/2=2

解题思路：灵活利用三角函数的公式进行化简，最后套“周期公式”。解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prced

sinx+cosx=1/3(sinx+cosx)²=sin²x+cos²x+2sinxcosx=1+sin(2x)=1/9sin(2x)=-8/9sinx+cosx=√2

记f(x)=sin2x+acos2x,则f(-π/8+x)=f(-π/8-x),代入化简,sin(-π/4+2x)+acos(-π/4+2x)=sin(-π/4-2x)+acos(-π/4-2x)当x

∵4sin2x-6sinx-cos2x+3cosx=0，∴4sin2x-cos2x-6sinx+3cosx=0，∴（2sinx+cosx）（2sinx-cosx）-3（2sinx-cosx）=0，∴（

ab=√3(sin2x)^2+cos2xsin2x=（√3/2）（1-cos4x）+（1/2）sin4x=（√3/2）+sin4xcos兀/3-cos4xsin兀/3=(√3/2)+sin(4x-兀/

sin2x/cosx=6/5,2sinx=6/5,sinx=3/5cos2x=1-2sin^2x=1-18/25=7/25

2cos2x

sin(2x)=2sinxcosx=2sinxcosx/(sin²x+cos²x)=2tanx/(tan²x+1)【同除以cos²x】=4/5

原式=f[cos(π/2-x)]=2-sin[2(π/2-x)]=2-sin(π-2x)=2-sin2x选C

f(x)=sin2x+mcos2xf(0)=mf(-π/3)=-gen3/2-m/2函数y=sin2x+mcos2x的图像关于直线x=-∏/6对称f(0)=f(-π/3)m=-gen3/3

∵向量a‖向量b所(2-sin2x)/(2+sin2x)=m/1m=(2-sin2x-2+2)/(2+sin2x)=[4-(sin2x+2)]/(2+sin2x)=[4/(2+sin2x)]-1因为-

1、sin2x=2sinxcosx而cos(x+45°)=cosx*cos45°-sinxsin45°=√2/2*(cosx-sinx)=0.6所以：cosx-sinx=3√2/5所以：（cosx-s

先用辅助角公式：设cosB=1/√(1+a^2),sinB=a/√(1+a^2),于是f(x)=√(1+a^2)·sin(2x+B)∵图像关于直线x=-π/6对称,∴x=-π/6时函数取到最值,∴2(

原式=(sin²x+cos²x+2sinxcosx)/(cos²x-sin²x)=(cosx+sinx)²/(cosx+sinx)(cosx-sinx

亲,记得给好评哦

求函数周期：Y=|sin2x|,y=|sin2x|+1/2Y=|sin2x|与y=|sin2x|+1/2的周期相同.y=sin2x与y=sin2x+1/2的最小正周期是π,那么Y=|sin2x|与y=

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