已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:
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已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)
![已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:](/uploads/image/z/757770-42-0.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CF%80-%CE%B8%29-cos%28%CF%80%2B%CE%B8%29%3D%E2%88%9A2%2F3%2C%E5%85%B6%E4%B8%AD%CF%80%2F2%EF%BC%9C%CE%B8%EF%BC%9C%CF%80%2C%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E7%9A%84%E5%80%BC%EF%BC%9A)
已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)
(1)sin(π-θ)-cos(π+θ)=√2/3
sinθ+cosθ=√2/3
(sinθ+cosθ)²=2/9
1+2sinθcosθ=2/9
2sinθcosθ=-7/9
sinθcosθ=-7/18
(sinθ-cosθ)²=1-2sinθcosθ=23/9
因为π/2<θ<π,所以sinθ-cosθ>0,所以sinθ-cosθ=√23/3
(2)sin³(π/2-θ)+cos³(π/2+θ)
=cos³θ-sin³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=√23/3*(1-7/18)
=11√23/54
打得好累啊
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)
(1)sin(π-θ)-cos(π+θ)=√2/3
sinθ+cosθ=√2/3
(sinθ+cosθ)²=2/9
1+2sinθcosθ=2/9
2sinθcosθ=-7/9
sinθcosθ=-7/18
(sinθ-cosθ)²=1-2sinθcosθ=23/9
因为π/2<θ<π,所以sinθ-cosθ>0,所以sinθ-cosθ=√23/3
(2)sin³(π/2-θ)+cos³(π/2+θ)
=cos³θ-sin³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=√23/3*(1-7/18)
=11√23/54
打得好累啊
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