2cos2x-2sin(x 3 2π)cos

①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c

y=2(1-2sin²x)+sin²x-4cosx=2-3sin²x-4cosx=2+3cos²-3-4cosx=3cos²-4cosx-1;再问：-

f(x)=(√3/2)sin2x+(1/2)cos2x+(√3/2)sin2x-(1/2)cos2x+cos2x+1=√3sin2x+cos2x+1=2sin(2x+π/6)+1周期T=2π/2=π对

诱导公式sin（a+π/2）=cosa所以sin（2x+π/2）=cos2x

（1）∵f(x)＝sin(2x+π6)+sin(2x−π6)+2cos2x=sin2xcosπ6+cos2xsinπ6+sin2xcosπ6−cos2xsinπ6+2cos2x+1=3sin2x+co

f(x)＝2sin2xcosπ3+cos2x＝sin2x+cos2x=2sin(2x+π4)…（4分）（1）函数f（x）的最小正周期为2π2＝π…（6分）（2）由题意知g（x）=f（x+3π8）+2=

(1+cos2x)²-2cos2x-1=1+2cos(2x)+cos²(2x)-2cos(2x)-1=cos²(2x)=[cos²(x)-sin²(x

看图片吧

2cos2x+sin^2x=2(cos^2x-sin^2x)+sin^2x=2cos^2x-sin^2x=3cos^2x-1

cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)]=[cos(n+1)x/2][cos((n-1)x/2

（Ⅰ）f(x)＝sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f（x）的最小正周期T＝2π2＝π（Ⅱ）当x∈[−π8，3π8]时，

即cos^8x-sin^8x=(cos^4x+sin^4x)(cos^4x-sin^4x)=(cos^4x+sin^4x)(cos²x+sin²x)(cos²x-sin&

∵f（x）=sin（π3−2x）+cos2x=32cos2x-12sin2x+cos2x=（32+1）cos2x-12sin2x=2+3sin（2x+θ）∴T=2π2=π故答案为：π．

sinx(cos^22x-sin^22x)+2cosxcos2xsin2x=sinxcos4x+cosxsin4x=sin(x+4x)=sin5x

f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))=(cos2x)^2/((sinπ/4cosx+cosπ/4sinx)(sinπ/4cosx-cos

因为sin^2(X)+cos^2(X)=1所以原式=1-cos^2（x）-cos^2（x）=1-2cos^2(x)=-(2cos^2(x)-1)=-cos2x

sin(x/2)的周期是4pi,cos2x的周期是pi,sin(x/2)+cos2x的周期是其最小公倍数,自然是4pi

f(sinx-1)=cos2x+2cos2x=1-2sin^2xf(sinx-1)=3-2sin^2x=-2(sinx-3/4)^2+7.5f(x)=-2(x+1/4)^2+7.5

sin2x+sin(π-x)/2cos2x+2sin^2x+cos(-x)=sin2x+sin(x)/2cos2x+2sin^2x+cos(x)=(2cosxsinx+sinx)/2cos²