赞 --个人流浪 幼苗
共回答了17个问题采纳率：82.4% 举报
(2cos^2a-1)/(2tan(pai/4+a)sin^2(pai/4-a))
= (cos2a)/{[2sin(π/4+a)/cos(π/4+a)]*sin²(π/4-a)}
∵ sin(π/4+a)=sin[(π/2)-(π/4-a)]=sin(π/4-a)
cos(π/4+a)=cos[(π/2)-(π/4-a)]=sin(π/4-a)
= (cos2a)/{[2cos(π/4-a)/sin(π/4-a)]*sin²(π/4-a)}
=(cos2a)/[2cos(π/4-a)*sin(π/4-a)]
=cos2a/sin(π/2-2a)
=cos2a/(cos2a)
=1
= (cos2a)/{[2sin(π/4+a)/cos(π/4+a)]*sin²(π/4-a)}
∵ sin(π/4+a)=sin[(π/2)-(π/4-a)]=sin(π/4-a)
cos(π/4+a)=cos[(π/2)-(π/4-a)]=sin(π/4-a)
= (cos2a)/{[2cos(π/4-a)/sin(π/4-a)]*sin²(π/4-a)}
=(cos2a)/[2cos(π/4-a)*sin(π/4-a)]
=cos2a/sin(π/2-2a)
=cos2a/(cos2a)
=1
1年前
8
共回答了1627个问题 举报
(2(cosa)^2-1)/[2tan(π/4+a)(sin(π/4-a))^2]
= cos2a. cos(π/4+a)/ [ 2( sin(π/4+a) (sin(π/4-a))^2 ]
= cos2a. cos(π/4+a)/ { (sin(π/4-a) [cos2a-cos(π/2) ] }
=cos(π/4+a)/ (sin(π/4-a)
=sin(π/4-a)/(sin(π/4-a)
=1
= cos2a. cos(π/4+a)/ [ 2( sin(π/4+a) (sin(π/4-a))^2 ]
= cos2a. cos(π/4+a)/ { (sin(π/4-a) [cos2a-cos(π/2) ] }
=cos(π/4+a)/ (sin(π/4-a)
=sin(π/4-a)/(sin(π/4-a)
=1
1年前
0
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