化简 sin(a b)cosx-1 2[sin(2a b)]

tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su

(1-cosx)'=sinx[(1-cosx)^2]'=(1-cosx)'*2(1-cosx)=2sinx(1-cosx)[sin(1-cosx)^2]'=[(1-cosx)^2]'*cos(1-co

sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)=sin^3x/cosx+cos^3x/sinx+2sinxcosx-(1+cosx/sinxcos

[sin^2(x)]/(sinx-cosx)-(sinx+cosx)/[tan^2(x)-1]-sinx=sin^2x/(sinx-cosx)-(sinx+cosx)/[(tanx+1)(tanx-1

等一下,你确定你那个b向量没写错?

左边=sin²x/(sinx-cosx)-(sinx+cosx)/(sin²x/cos²x-1)=sin²x(sinx+cosx)/(sinx-cosx)(si

cos（X-30°）+sin(X+30°)+cosX+1=cosX*cos30°+sinX*sin30°+sinX*cos30°+cosX*sin30°+cosX+1=[(根号3+1)/2]sinX+

1.y=√2sin(x+π/4),所以值域是[-√2,√2]2.y=1-cos^2x-cosx+1=-cos^2x-cosx+2设t=cosx,y=-t^2-t+2(=-1

本题应为：（sinx/cosx）*[√（1/sin²x）]-1否则因为sin²x-1小于0,如果是“根号1（/sin²x-1）”将无意义下面化简本题因为x为第二象限角,所

[1+2cosx/2*sinx/2+cos(x/2)^2-sin（x/2)^2][sin（x/2)-cos(x/2)]/根号下的[2*2cos(x/2)^2]=2cos(x/2)[sin（x/2)+c

因为sin(π/6)=1/2；cos(π/6)=(√3)/2所以原式等于cos(π/6)sinx+sin(π/6)cosx利用两角和公式sin(A+B)=sinAcosB+cosAsinB得到sin（

F(X)=2（COSX）^2+2*√3sinxcosx+1=-[1-2（COSX）^2]+√3sin(2x)+2=√3sin(2x)-[1-2（COSX）^2]+2=√3sin(2x)-cos(2x)

sin^2x/(sinx-cosx)-(sinx+cosx)/(tan^2x-1)=sin^2x/(sinx-cosx)-(sinx+cosx)/[(tanx+1)(tanx-1)]=sin^2x/(

原式=(1-cos^2x)/(1+cosx)-(1-sin^2x)/(1+sinx)=(1-cosx)(1+cosx)/(1+cosx)-(1-sinx)(1+sinx)/(1+sinx)=(1-co

sin⁴x/(sinx-cosx)-(sinx+cosx)cos²x/(tan²x-1)=sin⁴x/(sinx-cosx)-(sinx+cosx)cos&

sin(x-π/3)-cos(x+π/6)+√3cosx=sinxcosπ/3-cosxsinπ/3-cosxcosπ/6+sinxsinπ/6+√3cosx=1/2*sinx-√3/2*cosx-√

=2(cos(x)*cos(x)-1/2)^2/(2tan(π/4-x)cos(π/4-x)^2)=(1/2)*cos^2(2x)/[2sin(π/4-x)*cos(π/4-x)]=(1/2)*cos

sin2atana=2sinacosa*sina/cosa=2sin²a√(1-cosx)=√[2sin²(x/2)]=√2|sin(x/2)|sinπ/10×cosπ/5=cos

向量b＝（2cos,sinx)f(x)＝2cos^2x＋sinxcosx+1=1/2sin2x+cos2x+2=√5/2sin（2x+Φ）＋2这个函数的周期是π