2sin(x-a) 2*cos(x a) 2

根据余弦2倍角公式cos（a+b）=cosa*cosb-sina*sinbcos2a=cos（a+a）=cosa*cosa-sina*sina=cos²a-sin²a再根据三角函数

a=(1,-2sinx)b=(cos2x,cosx)f(x)=cos2x-2sinxcosx+a=cos2x-sin2x+a=√2cos(2x+π/4)+a(1).T=2π/2=π(2)0≤X≤π/2

cos平方x+cos平方(x+a)-2cosa*cosx*cos(x+a)=[cos2x+cos2(x+a)]/2+1-2cosa*cosx*cos(x+a)=cos(2x+a)cosa+1-2cos

f(x)=2sin(2wx-π/6)+λw=1/3+k/2(k=0,+_1.)w=5/6f(x)=2sin(5/3x-π/6)+λ(1)T=6π/5(2)f(x)=2sin(5/3x-π/6)-根号2

1,sin(pai/4-x)sin(pai/4+x)=cos(pai/4+x)sin(pai/4+x)=sin(pai/2+2x)/2=cos(2x)/22,cosA+cos(120度-A)+cos(

原式=-∫cos²xdcosx=-cos³x/3+C再问：第一步能讲一下为什么吗？再答：dcosx=-sinxdx采纳吧

sin^2x+cos^2x=1

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

sqrt(a^2+b^2)*sin(x+arctan(b/a))不是a/

cos^2x+cos(x+a)(cos(x+a)-2cosxcosa)=cos^2x+cos(x+a)(cosxcosa-sinxsina-2cosxcosa)=cos^2x+cos(x+a)(-co

=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答：别忘了负号再问：嗯谢谢

左边=（cosx)^2+[cos(x+a)]^2-[cos(x+a)+cos(x-a)]cos(x+a)=(cosx)^2-cos(x-a)cos(x+a)=(1/2)(1+cos2x-cos2x-c

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

sina-2cosa=0,则tana=2,则sina=2/根号5,-2/根号5cosa=1/根号5,-1/根号5则cos^2=1/54sin^2a-3sinaxcosa-5cos^2a=1/cos^2

∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x

2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

依题有2sin2x=sinθ+cosθsinx的平方=sinθ*cosθ又2sin2x=4sinx*cosxsinθ*cosθ=[(sinθ+cosθ)的平方-1]/2所以有sinx的平方=[(4si

f(x)=sin(a-3π)cos(2π-a)sin(-a+3π/2)/cos(-π-a)sin(-π-a)=sin(4π+a-3π)cosasin(3π/2-a)/cos(π+a)sin(2π-π-