
数学++--**//**--**/*/*/*/*/*/*--+-+--+-+-**--+-*-*
已知sin(3π)=1/3求〔sin(180°+α)cos(720°+α)tan(540°+α)*sin(-180°+α)〕/〔tan(900°+α)*sin(-180°...
已知sin(3π)=1/3求
〔sin(180°+α)cos(720°+α)tan(540°+α)*sin(-180°+α)〕/〔tan(900°+α)*sin(-180°-α)*cos(-180°-α)〕 展开
〔sin(180°+α)cos(720°+α)tan(540°+α)*sin(-180°+α)〕/〔tan(900°+α)*sin(-180°-α)*cos(-180°-α)〕 展开
2个回答
展开全部
〔sin(180°+α)cos(720°+α)tan(540°+α)*sin(-180°+α)〕/〔tan(900°+α)*sin(-180°-α)*cos(-180°-α)〕
= (- sina)*cosa*(-tana)*(-sina) / (-tana)*sina*(-cosa)
= -sina
sin(3π)=1/3 ???? sin(3π)= 0
= (- sina)*cosa*(-tana)*(-sina) / (-tana)*sina*(-cosa)
= -sina
sin(3π)=1/3 ???? sin(3π)= 0
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询