求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/22 09:20:59
求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1求证：2(sin2a+1)/(1+sin2a+cos2a
求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1
求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1
求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1
sin2a=2sinacosa 1=(sina)^2+(cosa)^2
2(sin2a+1)=2(sina+cosa)^2
1+sina2a+cos2a=1+2sinacosa+2(cosa)^2-1=2cosa(sina+cosa)
2(sin2a+1)/(1+sin2a+cos2a)=[2(sina+cosa)^2]/[2cosa(sina+cosa)]
=(sina+cosa)/cosa=tana+1
2(sin2a+1)/(1+sin2a+cos2a)
=2（2sinacosa+sin^2 a+cos^2 a)/(1+2sinacosa+2cos^2 a-1)
=2(sina+cosa)^2/2cosa(sina+cosa)
=(sina+cosa)/cosa
=tana+1
求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]
求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1
求证2sina+sin2a=2sin^3a/(1-cosa)
tana=2,则2sin2a+1/sin2a=2sin平方a+1/sin2a
（sin2a-cos2a)^2=1-sin4a 求证!
求证(cosa-sina)^2=1-sin2a
若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.
若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0
化简(2sin^2a+sin2a)/(1+tana)答案为sin2a
求证题能不能写详细点要步骤.急!（cos^2a-sin^2a)/sinacosa=2ctg2a、ctga-tga=2ctg2a、(1+sin2a-cos2a)/(1+sin2a+cos2a)=tga、2sin^2(a/2)+cosa=1
已知a、b为锐角,且满足3sin^2a+2sin^2b=1,3sin2a-2sin2b.求证a+2b=3.14/23sin2a-2sin2b=0
3sin^2A +2sin^2B=1,3sin2A- 2sin2B=0,求证A+2B=90度
求证(1+2sin2a/sin2a+1+cos2a)-(1/2)=(1/2)tana
求证（sin2a+1）/（1+cos2a+sin2a）=1/2tana+1/2
2sin^2a+sin2a/1-tana化简
tga=1/2,sin2a sin^2a=?
1+tana/2sin方a+sin2a化简帮帮忙
已知（2sin^a+sin2a)/(1+tana)=k (0