Question

已知tan(π/4+α)=2,求tanα和sin2α+sinα+cos2α的值

Answer 1

tan(π/4+α)=(tanα+1)/(1-tanα)=2 解得tanα=1/3所以sinα=3根号10/10 cosα=根号10/10 或 sinα= -3根号10/10 cosα= -根号10/10
sin2α+sinα+cos2α=2sinαcosα+sinα+1-2sinα的平方= （3根号10-2）/10或（3根号10+2）/10

Answer 2

由tan(π/4+α)=(tan(π/4)+tanα)/(1-tan(π/4)tanα)=(1+tanα)/(1-tanα)=2,
得tanα=1/3，
再由 tanα=sinα/cosα=1/3，(sinα)^2+(cosα)^2=1，得出sinα=√10/10,cosα=(3(√10))/10,
则sin2α+sinα+cos2α=2sinαcosα+sinα+1-2sin^2(α)=7/5+(√10)/10

Answer 3

tana=tan(a+π/4-π/4)
=[tan(a+π/4)-tan(π/4)]/[1+tan(a+π/4)tan(π/4)]
=(2-1)/(1+2*1)
=1/3
sin2a+sina+cos2a
=(2sinacosa+cos^2a-sin^2a)/(sin^2a+cos^2a)+sina
=(2tana+1-tan^2a)/(tan^2a+1)+sina
=(2/3+1-1/9)/(1/9+1)+sina
=7/5±√10/10

Answer 4

Answer 5