cos^2 80°+sin^2 50°-sin190°cos320°要求值._数学解答

cos^2 80°+sin^2 50°-sin190°cos320°
要求值.

人气：381 ℃　时间：2020-10-01 06:09:16

解答

sin^2 80 + sin^2 40 - cos50cos10
=cos^2 10 + sin^2 40 - cos50cos10
=(1/2)[(1+cos20)+(1-cos80)-(cos60+cos40)]
=(1/2)[3/2+cos20-(cos80+cos40)]
=(1/2)[3/2+cos20-2cos60cos20]
=3/4