sinθ+cosθ=4/3(0<θ<π/2),求sinθ-cosθ的值

发布网友发布时间：2024-10-08 20:57

共5个回答

热心网友时间：2024-12-03 19:36

可知（sin+cos)^2=16/9 sin^2+cos^2=1 所以2sin cos=7/9 所以 sin^2-2sin cos +cos^2=2/9 所以 sin-cos=根号2/3

热心网友时间：2024-12-03 19:36

因为sinθ+cosθ=4/3，(sinθ+cosθ)^2=16/9
sin^2θ+cos^2θ+2sinθcosθ=16/9,
2sinθcosθ=16/9-1=7/9
(sinθ-cosθ)^2=sin^2θ+cos^2θ-2sinθcosθ=1-7/9=2/9
sinθ-cosθ=±√2/3
希望对你有帮助

热心网友时间：2024-12-03 19:37

解：
sinθ+cosθ=4/3 ==> (sinθ+cosθ)² = 16/9
==> 1 + 2sinθ*cosθ =16/9
==> -(1-2sinθ*cosθ) = 16/9 -2
==> (sinθ - cosθ)² = 2/9
==> sinθ - cosθ = ±√2/3

热心网友时间：2024-12-03 19:37

sinθ+cosθ=4/3
(sinθ+cosθ)^2=(4/3)^2
1+2sinθcosθ=16/9
2sinθcosθ=7/9
(sinθ-cosθ)^2
=1-2sinθcosθ
=1-14/9
=5/9
sinθ-cosθ=±√5/3

热心网友时间：2024-12-03 19:38

sinθ+cosθ=4/3
(sinθ+cosθ)^2=16/9
sin^2θ+cos^2θ+2sinθcosθ = 16/9
1+2sinθcosθ = 16/9
2sinθcosθ= 7/9
(sinθ-cosθ)^2 = sin^2θ+cos^2θ-2sinθcosθ= 1-2sinθcosθ = 1-7/9 = 2/9
sinθ-cosθ = ±√2/3

sinθ+cosθ=4/3(0&lt;θ&lt;π/2),求sinθ-cosθ的值

sinθ+cosθ=4/3(0<θ<π/2),求sinθ-cosθ的值