y=cos(x π 3)-sin(x π 3)的最大值最小值周期

因为由上式可知y=cos(3x+π\4)=-sin[3(x-π/12)],要将y=-sin[3(x-π/12)]变换到y=sin（-3x),则需要作加法,即：-sin[3(x-π/12+π/12)],

y=sinxcos30+cosxsin30-cosxsin60-sinxcos60=sinx[(根号3-1)/2]+cosx[(1-根号3)/2]=[(根号3-1)/2](sinx-cosx)=[(根

y=cos(x+π/2）+sin（π/3-x)y=-sinx+sinπ/3cosx-sinxcosπ/3=-sinx+√3cosx/2-sinx/2=√3cosx/2-3sinx/2=√3(cosx/

COS（X+Y）COS（X-Y）=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

y=cos^2(3x+π/6)-sin^2(3x+π/6)=cos[2(3x+π/6)]=cos(6x+π/3)

如果说化简应该不对结果应该是常数+sinT或者cosT你的结果还能继续化下去

y＝（√3/2）sin（x＋π/2）＋cos（π/6－x）＝（√3/2）cosx＋cos（π/6）cosx＋sin（π/6）sinx＝（√3/2）cosx＋（√3/2）cosx＋（1/2）sinx＝√

再答：希望采纳！谢谢

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

再问：�

y'=2cos(2x-π/4)-3sin(3x+π/3)希望可以帮到你,如果解决了问题,请点下面的"选为满意回答"按钮,

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

y=cos(x+π/2)+sin(π/3-x)=-sinx+sinπ/3cosx-cosπ/3sinx=-sinx+√3/2cosx-1/2sinx=√3/2cosx-3/2sinx=√3(1/2co

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k

1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/

y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)=1+2cos^2(x)+sin2x=2+sin2x+cos2x构造向量a=（sin2x,cos2x）,b=（1,1）a+b=（si

正在做啊再问：恩再答：cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)y=sin(x+π/3)cos(π/6-x)=sin(x+π/3)sin(π/3+x)=sin

y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2

上下同除以cos（3x+π/4)得：原式=【1-tan(3x+π/4）】/[1+tan(3x+π/4）]用tan的和分角公式,得原式=[1-tan3x-1-tan3x]/[1-tan3x+1+tan3