大家来帮帮我吧,数学题...我想了1个多小时了求(cos24)^2+(sin6)^2+(cos18)^2

来源:百度知道 编辑:UC知道 时间:2024/06/01 19:34:45
求(cos24)^2+(sin6)^2+(cos18)^2

sin18°=(根号5-1)/4(必须公式)
原式=(cos24)^2+1-(cos6)^2+(cos18)^2
=(1+cos48-1-cos12+1+cos36)/2+1
=(cos48-cos12+1+cos36)/2+1
=[cos(60-12)-cos12+cos36+1〕/2+1
=[1/2cos12+根号3/2sin12-cos12+cos36+1〕/2+1
=[根号3/2sin12-1/2cos12+cos36+1〕/2+1
=[sin(12-30)+cos36+1〕/2+1
=[-sin18+cos36+1〕/2+1
=[-sin18+(cos18)^2-(sin18)^2+1]/2+1
=[-sin18+1-2(sin18)^2 +1]/2+1
代入 sin18°=(根号5-1)/4,即可得原式=1.75