穿越火线tgp准星:sin50°(1+根号3×cot80°)
来源:百度文库 编辑:高考问答 时间:2024/05/28 10:51:54
过程!
解:sin50°(1+√3×cot80°)
=sin50[sin80+√3×cos80]/sin80
=2sin50{(1/2)cos80+(√3/2)sin80}/sin80
=2sin50cos50/sin80
=sin100/sin80
=sin80/sin80
=1
解:sin50°(1+√3×cot80°)
=sin50[sin80+√3×cos80]/sin80
=2sin50{(1/2)cos80+(√3/2)sin80}/sin80
=2sin50cos50/sin80
=sin100/sin80
=sin80/sin80
=1