Question

若tanα=1/3,则2cos^2(α)+sin2α的值多少? 急

Answer 1

tanα=1/3,且sin^2(α)+cos^2(α)=1
原式=(2cos^2(α)+2sinαcosα)/(sin^2(α)+cos^2(α))
分号上下同除以cos^2(α), =(2+2tanα)/(tan^2(α)+1)
将tanα=1/3代入,得 =(2+2/3)/(1/9+1)=(8/3)/(10/9)=12/5

Answer 2

(cosa)^2 = 1/[(tana)^2 + 1]
= 9/10
sin2a = 2tana*(cosa)^2
= 2(cosa)^2/3
= 3/5
所以 2cos^2(a)+sin2a = 9/5 + 3/5
= 12/5

Answer 3

tan2a=(2/3)/(8/9)=3/4
=>sin2a=3/5,cos2a=4/5
或sin2a=-3/5,cos2a=-4/5
2cos^2a+sin2a=1+cos2a+sin2a
所以2cos^2a+sin2a=12/5或-2/5

Answer 4

tana=1/3，3sina=cosa，原式可化为3/8cos^2(a)因为sin2a=2/3Cos^2(a)，cosa=3比根号10，故原式=12/5

Answer 5

解：
2cos^2α+sin2α
=2cos^2α-1+1+sin2α
=cos2α+sin2α+1
=(1-tan^2α)/(1+tan^2α)+2tanα/(1+tan^2α)+1
=(1+2tanα-tan^2α)/(1+tan^2α)+1
=(1+2/3-1/9)/(1+1/9)+1
=12/5

Answer 6

sinα/cosα=1/3 ,
似乎少条件