求函数y=cos(x-3)的最小正周期

解题思路：考查三角恒等变换解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/read

y=cosX+cos(X-π/3)=cosX+cosX*1/2+sinX*(√3)/2=cosX*3/2+sinX*(√3)/2=3sin(X+β)∴最小值：-3最大值：3

y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/

y=-2(cosx)^2+2sinx+3/2=(sinx)^2-2+2sinx+3/2设sinx=t=t^2+2t-1/2=(t+1)^2-3/2此y-t函数形状为关于-1对称的正二次曲线所以t=-1

y=sec²x+2tanx+1=tan²x+2tanx+2=(tanx+1)²+1x∈[-π/3,π/4],则tanx∈[-√3,1]当tanx=-1,即x=-π/4时,

可以根据：asinα+bcosα=√(a^2+b^2)*sin(α+θ),其中θ由a,b的符号和tanθ=b/a确定具体到这题,就是：y=√[(√3+2)/2]^2+(1/2)^2*sin(2x+θ)

x=-pi/6,y=-1,最小,x=pi/6,y=5,最大再问：什么意思看不懂再答：pi就是圆周率的那个π再问：没过程a，为啥要取这两个值我不懂

y=-2cos^2x+2sinx+3/2=-2+2sin^2x+2sinx+3/2=-2+2(sin^2x+sinx+1/4)+1=2(sinx+1/2)^2-1当sinx=-1/2即x=-π/6时,

解y=5cos²x+3sin²x-2sinxcosx=3cos²x+3sin²x+2cos²x-2sinxcosx=3+2cos²x-2si

y=sin^2x+2sinxcosx+3cos^2x=sin^2x+2sinxcosx+cos^2x+2cos^2x=1+sin2x+2cos^2x-1+1=sin2x+cos2x+2=√2(√2/2

y=cos^2x-sin^2-√3cos(3π/2+2x)+1=cos2x-√3sin2x+1=2cos(2x+π/3)+1当2kπ+π/2≤2x+π/3≤2kπ+3π/2时,函数单调递减2kπ+π/

y=3cosX-cos(2X)=3cosx-(2*(cosx^)2-1)=-2(cosx)^2+3cosx+1=-2(cosx-3/4)^2+17/8当cos=3/4时,y有最大值,为17/8当cos

y'=-sin(4-3X)*(-3)=3sin(4-3X)

和差化积得y=2cos(x-π/6)cosπ/6=√3cos(x-π/6).故ymin=-√3,ymax=√3.

y=（1/cosx）+2tanx+1=secx+2tanx+1=tanx+2tanx+2=(tanx+1)+1由：x∈[-π/3,π/4]→1-√3≤tanx+1≤2故：当x=-π/4时,ymin=0

y=cos²x－sinxcosx=1/2(1+cos2x)-1/2sin2x=1/2cos2x-1/2sin2x+1/2=√2/2(√2/2*cos2x-√2/2*sin2x)+1/2=√2

y=cosx*(-2)sin[(x+x+2/3)/2]sin[(-2/3)/2]=2cosxsin(x+1/3)sin(1/3)=2sin(1/3)cosxsin(x+1/3)=2sin(1/3)(1

y=sin^2x+3cos^2x=sin^2x+cos^2x+2cos^2x=1+1+cos2x=2+cos2xT=2π/2=π取最大值x=2kπ取最小值时x=2kπ+π

X∈(0,3/π）x-π/6∈[-π/6,π/6]y=cos[2(x-π/6)]+2sin(x-π/6)=1-2sin^2(x-π/6)+2sin(x-π/6)令sin(x-π/6)=t(-1/2≤t

y=2asinx-cos²x+a²+2=2asinx+sin²x+a²+1=(sinx+a)²+1当a>=0时最小值为f(x)=(a-1)²