(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/29 13:33:49
(救命阿!)
证明下列恒等式.
2(sin2a+1)
-----------------=1+tana
1+sin2a+cos2a
2
-------------=tana+cota
sin2a
2sina +sin2a=2sin3立方a
-----------
1-cosa
已知3sinB=sin(2a+B),求证:tan(a+B)=2tana
第二题打错了.
应该是.
2
---------=2sin立方a/1-cosa
sin2a
证明下列恒等式.
2(sin2a+1)
-----------------=1+tana
1+sin2a+cos2a
2
-------------=tana+cota
sin2a
2sina +sin2a=2sin3立方a
-----------
1-cosa
已知3sinB=sin(2a+B),求证:tan(a+B)=2tana
第二题打错了.
应该是.
2
---------=2sin立方a/1-cosa
sin2a
1)
2(sin2α+1)
=2*2sinαcosα+2
=4sinαcosα+2
1+sin2a+cos2a
=1+2sinαcosα+2cos^2(α)-1
=2sinαcosα+2cos^2(α)
2(sin2α+1)/(1+sin2a+cos2a)
=(4sinαcosα+2)/(2sinαcosα+2cos^2(α))=(2sinαcosα+1)/(sinαcosα+cos^2(α))
=[(sinαcosα+cos^2(α))+(sinαcosα-cos^2(α)+1)]/sinαcosα+cos^2(α))
=1+(sinαcosα-cos^2(α)+1)/(sinαcosα+cos^2(α))
=1+(sinαcosα+sin^2(α))/(sinαcosα+cos^2(α))
=1+sinα(cosα+sinα)/[cosα(cosα+sinα)]
=1+sinα/cosα
=1+tanα
2)
2/sin2α
=1/(sinαcosα)
=[sin^(α)+cos^2(α)]/(sinαcosα)
=sinα/cosα+cosα/sinα
=tanα+cotα
3)
2sin^3(α)/(1-cosα)
=2sin^2(α)sinα/(1-cosα)
=2(1-cos^2(α))sinα/(1-cosα)
=2(1+cosα)sinα
=2sinα+2cosαsinα
=2sinα+sin2α
4)
3sinB=sin(2a+B)
→3sin[(a+B)-a]=sin[(a+B)+a]
sin(α±β)=sinα·cosβ±cosα·sinβ
所以,
3sin(a+B)cosa-3cos(a+B)sina=sin(a+B)cosa+cos(a+B)sina
得到
2sin(a+B)cosa=4cos(a+B)sina
sin(a+B)cosa=2cos(a+B)sina
两边同除cos(a+B)cosa
tan(a+B)=2tana
2(sin2α+1)
=2*2sinαcosα+2
=4sinαcosα+2
1+sin2a+cos2a
=1+2sinαcosα+2cos^2(α)-1
=2sinαcosα+2cos^2(α)
2(sin2α+1)/(1+sin2a+cos2a)
=(4sinαcosα+2)/(2sinαcosα+2cos^2(α))=(2sinαcosα+1)/(sinαcosα+cos^2(α))
=[(sinαcosα+cos^2(α))+(sinαcosα-cos^2(α)+1)]/sinαcosα+cos^2(α))
=1+(sinαcosα-cos^2(α)+1)/(sinαcosα+cos^2(α))
=1+(sinαcosα+sin^2(α))/(sinαcosα+cos^2(α))
=1+sinα(cosα+sinα)/[cosα(cosα+sinα)]
=1+sinα/cosα
=1+tanα
2)
2/sin2α
=1/(sinαcosα)
=[sin^(α)+cos^2(α)]/(sinαcosα)
=sinα/cosα+cosα/sinα
=tanα+cotα
3)
2sin^3(α)/(1-cosα)
=2sin^2(α)sinα/(1-cosα)
=2(1-cos^2(α))sinα/(1-cosα)
=2(1+cosα)sinα
=2sinα+2cosαsinα
=2sinα+sin2α
4)
3sinB=sin(2a+B)
→3sin[(a+B)-a]=sin[(a+B)+a]
sin(α±β)=sinα·cosβ±cosα·sinβ
所以,
3sin(a+B)cosa-3cos(a+B)sina=sin(a+B)cosa+cos(a+B)sina
得到
2sin(a+B)cosa=4cos(a+B)sina
sin(a+B)cosa=2cos(a+B)sina
两边同除cos(a+B)cosa
tan(a+B)=2tana
(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2
证明 2(sin2a+1)/1+sin2a+cos2a=tana+1
证明:(sin2a+1)/(sin2a+cos2a+1)=1/2(tana+1)
求证:2(sin2a+1)/(1+sin2a+cos2a)=tana+1
(sina-cosa)^2=1-sin2a怎么证明,
(sin2a-cos2a)^2=1-sin4a 求证!
(sin2a-cos2a)^2=1-sin4a
在△ABC中,求证:(1)sin2A+sin2B+sin2C=2+2cosAcosBcosC;(2)cos2A+cos2
求证:1+sin2a-cos2a/1+sin2a+cos2a=tana
求证1+sin2a-cos2a除以1+sin2a+cos2a=tana
化简:(1+cos2a)/sin2a-sin2a/(1-cos2a)=
证明sin2a/(1+sina+cosa)=sina+cosa-1