已知tan(α/2)=2 求sin(α+π/6)的值

问题描述:

已知tan(α/2)=2 求sin(α+π/6)的值

tan(α/2)=2
sinα=2tan(α/2) /[1+tan²(α/2)]
=2×2/(1+4)
=4/5
cosα=[1-tan²(α/2)]/[1+tan²(α/2)]
=(1-4)/(1+4)
= -3/5
sin(α+π/6)=sinαcosπ/6+cosαsinπ/6
=4/5×√3/2-3/5×1/2
=(4√3-3)/10

令a=α/2则tana=2sinα=sin2a=2sinacosa=2sinacosa/(sin²a+cos²a)上下除以cos²a=2tana/(tan²a+1)=4/5cosα=cos2a=cos²a-sin²a=(cos²a-sin²a)/(cos²a+sin²a)上下...