题目

（1）sin40度(ta n10度-根号下3) （2）tan70度cos10度（根号下3*tan20度-1） 这俩咋化简啊?

提问时间：2020-07-30

答案

(1) sin40(sin10/cos10 - sin60/Cos60)
=Sin40/(Cos10Cos60) (Sin 10 Cos60 - Cos10Sin60)
=Sin40 Sin (-50)/ (Cos10 Cos60)
=-Sin40 Cos40 / (Sin80 Cos 60)
= - Sin80/ 2 /(Sin80 Cos 60)
= - 1/(2Cos60)
=-1
(2)Sin70/Cos70 Cos10 * (Sin 60/Cos60 Sin20 /Cos20 -1)
= Cos20 Cos 10/Cos70 *(Sin60 Sin20 - Cos60 Cos20)/ (Cos60 Cos20)
= Cos10 /Cos70 * (-Cos40) / Cos60
= - Cos 10 Cos40 /(Cos70 Cos60)
= - 2 Sin80 Cos40/ Sin 20
= - 8 sin20 Cos 20 Cos40 /Sin20
= -8 Cos20 Cos40
=-4 (Cos 20 - Cos (60)]
= -4 Cos 20 +2