若α为第四象限角,且cosα=12/13,则tanα/2=

问题描述:

若α为第四象限角,且cosα=12/13,则tanα/2=

cosα= 1-2{sin(α/2)}^2
12/13= 1-2{sin(α/2)}^2
sin(α/2) = √(1/26)

cosα= 2{cos(α/2)}^2-1
12/13= 2{cos(α/2)}^2-1
cos(α/2) = -√(25/26)
tan(α//2) = sin(α/2)/cos(α/2)
= -1/5

a是第四象限角,即有270135cosa=(cos^2a/2-sin^2a/2)/(sin^2a/2+cos^2a/2)
=(1-tan^2 a/2)/(tan^2 a/2+1),(上下同除以cos^2a/2)
12/13=(1-tan^2a/2)/(tan^2a/2+1)
13(1-tan^2a/2)=12(tan^2a/2+1)
25tan^2a/2=1
tan^2a/2=1/25
故有:tana/2=-1/5

因为 α 是第四象限角,所以 sinα= -5/13 ,
则 tan(α/2)
=sin(α/2)/cos(α/2)
=2sin(α/2)cos(α/2)/[2(cosα/2)^2]
=sinα/(1+cosα)
=(-5/13)/(1+12/13)
= -1/5 .