cos2(π 4-a)

∵f（x）=cos2(x−π4)−cos2(x+π4)=1+cos(2x−π2)2-1+cos(2x+π2)2=sin2x2-−sin2x2=sin2x．∴T=2π2=π，又f（-x）=sin（-2x

tan(α+π/4)=(1+tanα)/(1-tanα)=(sinα+cosα)/(cosα-sinα)=(sinα+cosα)^2/(cosα^2-sinα^2)=(1+sin2α)/cos2α=(

COS^2(π/2+A)+COSA=5/4由COS^2(π/2+A)=[cos(π/2+A)]^2=[-sin(A)]^2=sin^2A=1-cos^2A则:1-cos^2A+cosA=5/4cos^

原式=sinπ7(cos2π7+cos4π7+cos6π7)sinπ7=sinπ7cos2π7+sinπ7cos4π7+sinπ7cos6π7sinπ7=12(sin3π7−sinπ7)+12(sin

f(x)=2(cosx)^2+2√3sinxcosx+a=√3sin2x+cos2x+1+a=2sin(2x+π/6)+1+a(1)f(π/6)=4f(π/6)=2sin(2*π/6+π/6)+1+a

sin(5/6π-a)+cos2(7／6π+a)=sin[π-(5/6π-a)]+1-2sin²(7/6π+a)=sin(π/6+a)+1-2sin²(π+π/6+a)]=1/3+

cos2(π/4-α)-sin2(π/4-α)=cos(π/2-2α）-sin(π/2-2α)=sin2α-cos2α=√2(√2/2sin2α-cos2α)=√2sin(2α-π/4)根据二倍角公式

由两角和公式展开第一个等式得：根号2（sina+cosa）=sinθ+cosθ,再两边平方,得2(1+sin2a)=1+sin2θ得sin2a=(sin2θ-1)/2;cos2β=1-2sin^2β=

2sin（π/4＋α）=√2(sina+cosa)√2(sina+cosa)=sinθ＋cosθ将这个式子平方,得2(1+sin2a)=1+sin2θ2sin2β=sin2θ2(1+sin2a)=1+

cos^2(π/2-a)+cos2^2(π/6+a)=sin^2a+1/2(1+cos(π/3+2a)=1/2(1-cos(π/3+2a)+1/2(1+cos(π/3+2a)=1再问：cos^2(3/

(1)a·b=(cos2/3x,sin2/3x)*(cos2/x,-sin2/x)=cos2/3x*cos2/x-sin2/3x*sin2/x=cos(2/3x+2/x)=cos8/3x|a+b|=√

1f(x)=a·b+2λ|a+b|a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))=cos(2x)|a+b|^2=|a|^2+|b|^2+2a·b=2+2c

由直角得：OA·OB=(a-b)(a+b)=a²-b²=0∴‖a‖=‖b‖由等腰得：‖OA‖=‖OB‖即‖a-b‖=‖a+b‖∴√（a-b）²=√（a+b）²∴

a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),(1)a*b=(cos3x/2,sin3x/2)*(cosx/2,-sinx/2)=cos(3x/2)*cos(x/2)-

原式=cos[2(π/4-a)]=sin(2a)公式：cos2x=(cosx)^2-(sinx)^2

sin（π/2+a）=cosa=-√5/5∴a∈(π/2,π)∴sina=2√5/5[cos²（π/4+a/2）-cos²（π/4-a/2）]/[sin（π-a）+cos（3π+a

令：a+π/3=bcos2b=1-2(sinb)^2=1-(1/3)^2=7/9

tan(α+π/4)=(1+tanα)/(1-tanα)=(sinα+cosα)/(cosα-sinα)=(sinα+cosα)^2/(cosα^2-sinα^2)=(1+sin2α)/cos2α=(

设A=α+π/4,2A=2α+π/2tan(2A)=tan(2α+π/2)=-cot(2α)=-cos2α/sin2α=-b/atan(2A)=2tanA/(1-tanA^2)=-b/a解之得[2a+

tan[π/4+α]=(tanπ/4+tanα)/(1-tanπ/4tanα)=(1+sinα/cosα)/(1-sinα/cosα)=(cosα+sinα)/(cosα-sinα)=[2cos^2(