Question

已知sinθ+cosθ=√2/3(π/2<θ<π),求下列各式的值

已知sinθ+cosθ=√2/3(π/2<θ<π),求下列各式的值
（1）sinθ-cosθ
（2）tanθ
（3）sin^(3)θ-cos^(3)θ

过程。步骤。谢谢！！！！

Answer 1

(1) π/2<θ<π

sinθ>0
cosθ<0
sinθ-cosθ>0
sinθ+cosθ=√2/3
(sinθ+cosθ)²=(√2/3)²

1+2sinθcosθ=2/9
sinθcosθ=-7/18
sinθ-cosθ=√(1-2sinθcosθ)
=√(1-2*(-7/18))
=√(32/18)
=4/3
(2) sinθ+cosθ=√2/3......(1)
sinθ-cosθ=4/3.......(2)
(1)+(2)：2sinθ=(4+√2)/3
sinθ=(4+√2)/6
(4+√2)/6-cosθ=4/3
cosθ=(4+√2-8)/6
=(√2-4)/6
tanθ=sinθ/cosθ
=((4+√2)/6)/((√2-4)/6)
=(4+√2)(√2+4)/(2-16)
=(16+2+8√2)/(-14)
=-9/7-4√2/7
(3) sin³θ-cos³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=4/3(1-7/18)
=4/3*11/18
=22/27

Answer 2

已知sinθ+cosθ=√2/3(π/2<θ<π),求下列各式的值
（1）
sinθ+cosθ=√2/3 (1)

sin^2θ+2sinθcosθ+cos^2θ=2/9

2sinθcosθ=-7/9
sin^2θ-2sinθcosθ+cos^2θ=2/9-4sinθcosθ
(sinθ-cosθ)^2=2/9-4*(-7/9)
(sinθ-cosθ)^2=30/9
π/2<θ<π

sinθ>cosθ

sinθ-cosθ=√30/3 (2)
2题
由(1)+(2)得
2sinθ=√2/3+√30/3
由(1)-(2)得
2cosθ=√2/3-√30/3
tanθ=(√2/3+√30/3)/(√2/3-√30/3)
tanθ=(√2+√30)/(√2-√30)
tanθ=(1+√15)/(1-√15)
tanθ=(1+2√15+15)/(1-15)
tanθ=-(8+√15)/7
（3）sin^(3)θ-cos^(3)θ

=(sinθ-cosθ)(sin^(2)θ+sinθcosθ+cos^(2)θ)
=(√30/3)(1-7/18)
=11√30/54