cos40(1+根号3*cot80)求值怎么从2cos40cos50/sin80 到 (cos10+cos90)/cos10=2cos40(1/2+根号3/2*cos80/sin80)=2cos40(sin80sin30+cos30cos80)/sin80=2cos40cos50/sin80=(cos10+cos90)/cos10=1
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/10 15:36:15
![cos40(1+根号3*cot80)求值怎么从2cos40cos50/sin80 到 (cos10+cos90)/cos10=2cos40(1/2+根号3/2*cos80/sin80)=2cos40(sin80sin30+cos30cos80)/sin80=2cos40cos50/sin80=(cos10+cos90)/cos10=1](/uploads/image/z/6878857-49-7.jpg?t=cos40%281%2B%E6%A0%B9%E5%8F%B73%2Acot80%29%E6%B1%82%E5%80%BC%E6%80%8E%E4%B9%88%E4%BB%8E2cos40cos50%2Fsin80+%E5%88%B0+%28cos10%2Bcos90%29%2Fcos10%3D2cos40%281%2F2%2B%E6%A0%B9%E5%8F%B73%2F2%2Acos80%2Fsin80%29%3D2cos40%28sin80sin30%2Bcos30cos80%29%2Fsin80%3D2cos40cos50%2Fsin80%3D%28cos10%2Bcos90%29%2Fcos10%3D1)
x)K/610~`Z%66=mΎ'~qfӎ
@Cm iiA7o4Q`
j4!l1fMR>= 6/P
B$!87/uۓku:l=t