2个回答
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解:
sin³θ+cos³θ = (sinθ+cosθ)(sin²θ-sinθ*cosθ+cos²θ)
= (sinθ+cosθ)*1/2*[3-(sin²θ+2sinθ*cosθ+cos²θ)]
= 1/2*(sinθ+cosθ)*[3- (sinθ+cosθ)²]=1
因此:
(sinθ+cosθ)³升铅 - 3(sinθ+cosθ)+2 =0
==> [(sinθ+cosθ)-1]² * [(sinθ+cosθ)+2]=0
∵|sinθ+cosθ|= √唤笑凯2|sin(θ+π/4)|≤ √2
∴(sinθ+cosθ)+2>0
因此: (sinθ+cosθ)-1=0 ==>和唤(sinθ+cosθ) =1
sin³θ+cos³θ = (sinθ+cosθ)(sin²θ-sinθ*cosθ+cos²θ)
= (sinθ+cosθ)*1/2*[3-(sin²θ+2sinθ*cosθ+cos²θ)]
= 1/2*(sinθ+cosθ)*[3- (sinθ+cosθ)²]=1
因此:
(sinθ+cosθ)³升铅 - 3(sinθ+cosθ)+2 =0
==> [(sinθ+cosθ)-1]² * [(sinθ+cosθ)+2]=0
∵|sinθ+cosθ|= √唤笑凯2|sin(θ+π/4)|≤ √2
∴(sinθ+cosθ)+2>0
因此: (sinθ+cosθ)-1=0 ==>和唤(sinθ+cosθ) =1
追问
= (sinθ+cosθ)*1/2*[3-(sin²θ+2sinθ*cosθ+cos²θ)]
这步里面的1/2*[3-(sin²θ+2sinθ*cosθ+cos²θ)]是怎么来的?
谢谢
追答
让你久等了;
sin²θ-sinθ*cosθ+cos²θ =1/2*(2sin²θ-2sinθ*cosθ+2cos²θ)
=1/2*[3(sin²θ+cos²θ)-(sin²θ+cos²θ)-2sinθ*cosθ]
=1/2*[3-(sin²θ+cos²θ+2sinθ*cosθ)]
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