Question

已知tanα=2,求(sinα-cos^5α)/(3sin^3α-cos^3α）

已知tanα=2,求(sinα-cos^5α)/(3sin^3α-cos^3α）

Answer 1

已知tanα=2
则(sinα-cos^5α)/(3sin^3α-cos^3α)
=(tanα/cos^2α-cos^2α)/(3tan^3α-1)
=(2/cos^2α-cos^2α)/23

而1/cos^2α=(sin^2α+cos^2α)/cos^2α=1+tan^2α=1+4=5
所以cos^2α=1/5

所以原式=(2/cos^2α-cos^2α)/23
=(2*5-1/5)/23
=49/115

Answer 2

因为tanα=2，所以cos^2α=1/5
(sinα-cos^5α)/(3sin^3α-cos^3α）
=(sinα/cos^3α-cos^2α)/(3tan^3α-1)
=(tanα/cos^2α-cos^2α)/23
=(2/(1/5)-1/5)/23
=49/115