(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2a2-------------=tana+cotasin2a2sina +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-c
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![(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2a2-------------=tana+cotasin2a2sina +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-c](/uploads/image/z/5308660-28-0.jpg?t=%28%E6%95%91%E5%91%BD%E9%98%BF%21%29%E8%AF%81%E6%98%8E%E4%B8%8B%E5%88%97%E6%81%92%E7%AD%89%E5%BC%8F.2%28sin2a%2B1%29-----------------%3D1%2Btana1%2Bsin2a%2Bcos2a2-------------%3Dtana%2Bcotasin2a2sina+%2Bsin2a%3D2sin3%E7%AB%8B%E6%96%B9a-----------1-cosa%E5%B7%B2%E7%9F%A53sinB%3Dsin%282a%2BB%29%2C%E6%B1%82%E8%AF%81%3Atan%28a%2BB%29%3D2tana%E7%AC%AC%E4%BA%8C%E9%A2%98%E6%89%93%E9%94%99%E4%BA%86.%E5%BA%94%E8%AF%A5%E6%98%AF.2---------%3D2sin%E7%AB%8B%E6%96%B9a%2F1-c)
(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2a2-------------=tana+cotasin2a2sina +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-c
(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
(救命阿!)证明下列恒等式.2(sin2a+1)-----------------=1+tana1+sin2a+cos2a2-------------=tana+cotasin2a2sina +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-c
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