(1)ä¾é¢æå¾
å 为,αâï¼0,Ï/2ï¼,tanα=1/2
æ以,tan2a=2tana/[1-(tana)^2]
=2*1/2/[1-1/4]
=1/3/4
=4/3
(2)å 为αâï¼0,Ï/2ï¼
æ以,sina>0,cosa>0 2aâï¼0,Ïï¼
å 为,tana=1/2,(sina)^2+(cosxï¼^2=1
解å¾sina=â5/5,cosa=2â5/5
æ以sin2a=2sinacosa=4/5,cos2a=±3/5
å½sin2a=4/5,cosa=3/5æ¶,
sin(2α+Ï/3)=1/2*sin2a+â3/2*cos2a
=1/2*4/5+â3/2*3/5
=(4+3â3)/10
å½sin2a=4/5,cosa=-3/5æ¶,
sin(2α+Ï/3)=1/2*sin2a+â3/2*cos2a
=1/2*4/5-â3/2*3/5
=(4-3â3)/10
sin(2α+Ï/3)
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