Question

设cosα=-√5/5 π<α<3π/2 tanβ=1/3 0<β<π/2，求（α-β） 谢谢啦！！

Answer 1

cosα=-√5/5
π＜α＜3π/2
∴sinα=-2√5/5
tanα=2
tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)
=[2-(1/3)]/[1+2*(1/3)]
=(5/3)/(5/3)
=1
0<β＜π/2
π/2＜α-β＜3π/2
α-β=5π/4

Answer 2

sinα=-2根号5/5
cosα=-根号5/5
tanβ=1/3
sinβ=根号10/10
cosβ=3根号10/10

sin(α-β)=sinαcosβ-cosαsinβ
=-6根号50/50+根号50/50
=-5根号50/50
=-根号2/2

α-β=5π/4