å·²ç¥x1,x2æ¯å½æ°f(x)=sin(wx+q)(w>0,0<q<Ï)çä»»æ两个ç¸é»é¶ç¹ä¸|x1-x2|=Ï ç¹(Ïï¼-1ï¼å¨å½æ°
f(x)çå¾è±¡ä¸ æ±ï¼1ãf(x)è¡¨è¾¾å¼ 2ãè¥aå±äº(0ï¼Ï/2)ä¸f(2a)=3/5 æ±1/(sina+cosa)çå¼
解ï¼Ïx₁+q=0ï¼æ
x₁=-q/Ï...........(1)
Ïx₂+q=Ïï¼æ
x₂=(Ï-q)/Ï............(2)
︱x₁-x₂︱=(Ï-q)/Ï+q/Ï=Ï/Ï=Ïï¼æ
Ï=1,
äºæ¯å¾f(x)=sin(x+q)ï¼åf(Ï)=sin(Ï+q)=-sinq=-1ï¼æ
q=Ï/2
â´f(x)ç表达å¼ä¸ºf(x)=sin(x+Ï/2)=cosx.
f(2α)=cos(2α)=2cos²Î±-1=3/5ï¼cos²Î±=4/5ï¼cosα=2/â5ï¼sin²Î±=1-4/5=1/5ï¼sinα=1/â5
â´1/(sina+cosa)=1/(1/â5+2/â5)=(â5)/3.
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