Question

已知tanθ=2 求sinθ/Sin³θ+cos³θ=？.

要过程

Answer 1

tanx=2
∵sin²x+cos²x=1
∴原式
= [sinx(sin²x+cos²x)]/(sin³x+cos³x)
=(sin³x+sinxcos²x)/(sin³x+cos³x)
=(tan³x+tanx)/(tan³x+1)
=(8+2)/(8+1)
=10/9

Answer 2

sina/[(sina)^3+(cosa)^3]
=(sina/cosa)/[(sina/cosa)*(sina)^2+(cosa)^2]
=tana/[tana*(1/(1+1/(tana)^2) +1/(1+(tana)^2)]
=2/[2*1/(1+1/4)+1/(1+4)]
=2/(8/5+1/5)
=10/9