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