已知sin2x=m.cos2x=n.求tan(x+π/4)的值

问题描述:

已知sin2x=m.cos2x=n.求tan(x+π/4)的值

tan(x+π/4)
=(tanx+1)/(1-tanx)
上下同乘cosx
=(sinx+cosx)/(cosx-sinx)
sin2x=2sinxcosx
1+2sinxcosx
=sin²x+cos²x+2sinxocsx
=(sinx+cosx)²
=1+m
cos2x
=cos²x-sin²x
=(cosx-sinx)(cosx+sinx)
=n
∴(sinx+cosx)²/[(cosx-sinx)(cosx+sinx)]
=(sinx+cosx)/(cosx-sinx)
=(1+m)/n
所以
tan(x+π/4)
=(1+m)/n