Question

已知tanα=根号3，π<α<3/2π，求sinα-cosα的值

Answer 1

π<α<3/2π
0<α-π<π/2
tan(α-π)=根号3
α-π=π/3
α=4π/3
sinα-cosα=-根号3/2-(-1/2)=(1-根号3)/2

追问

sinα-cosα=-根号3/2-(-1/2)=(1-根号3)/2 ?不是等于(1+根号3)/2 么？

追答

sin=4π/3=-根号3/2

Answer 2

∵tanα=√3, ∴α=π/3. α在第Ⅲ象限，
sinα-cosα=√2sin(α-π/4). 【π＜α＜3π/2 π-π/4<,α-π/4<3/2π-π/4. 3π/4＜α-π/4<5π/4.】
=√2sin(π/3-π/4).=
=√2sin(-π/12).
=-√2sinπ/12.
=√2/2(1-√3). ----即为所求。