(1)=______;(2)sin2l°+sin22°+…+sin288°+sin289°=______.

发布时间:2020-08-06 03:47:44

(1)=______;
(2)sin2l°+sin22°+…+sin288°+sin289°=______.

网友回答

解:(1)∵tan63°=cot27°,tan27°<tan63°,
∴tan27°<cot27°.

=
=
=|tan27°-cot27°|
=cot27°-tan27°;
sin2l°+sin22°+…+sin288°+sin289°
=(sin2l°+sin289°)+(sin22°+sin288°)+…+(sin244°+sin246°)+sin245°
=1+1+…+1+
=44.
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