已知sin2α=a,cos2α=b,则tan[(π/4+α]等于?

tan[π/4+α]
=(tanπ/4+tanα)/(1-tanπ/4tanα)=(1+sinα/cosα)/(1-sinα/cosα)
=(cosα+sinα)/(cosα-sinα)=[2cos^2(α)+2cosαsinα]/[2cos^2(α)-2cosαsinα]
=[1+cos2α+sin2α]/[1+cos2α-sin2α]=(1+a+b)/(1-a+b)
tan[π/4+α]
=(tanπ/4+tanα)/(1-tanπ/4tanα)=(1+sinα/cosα)/(1-sinα/cosα)
=(cosα+sinα)/(cosα-sinα)=[2cosαsinα+2sin^2(α)]/[2cosαsinα-sin^2(α)]
=[sin2α+1-cos2α]/[sin2α-1+cos2α]=(a+1-b)/(a-1+b)
tan[π/4+α]
=sin(π/4+α)/cos(π/4+α)=(sinα+cosα)/(cosα-sinα)
=(sinα+cosα)^2/[(cosα-sinα)(sinα+cosα)]=(1+2sinαcosα)/[cos^2(α)-sin^2(α)]
=(1+sin2α)/cos2α=(1+a)/b
tan[π/4+α]
=sin(π/4+α)/cos(π/4+α)=(sinα+cosα)/(cosα-sinα)
=[(sinα+cosα)(cosα-sinα)]/[(cosα-sinα)^2]=[cos^2(α)-sin^2(α)]/(1-2sinαcosα)
=cos2α/(1-sin2α)=b/(1-a)
这四个表达式相等,而且可以利用a^2+b^2=1来转换
例如：
(1+a+b)/(1-a+b)
=[(b+1+a)(b-1-a)]/[(-a+b+1)(-a+b-1)]=[b^2-(1+2a+a^2)]/[(a^2+b^2-2ab)-1]
=[-2a-2a^2]/[-2ab]=(1+a)/b