设cos(180°+α)=1/3 且 sinα>α 求sin(180°+α)-tanα/cos(180°-α)+tan(
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设cos(180°+α)=1/3 且 sinα>α 求sin(180°+α)-tanα/cos(180°-α)+tan(180°-α)
cos(180°+α)=1/3,则cosa=-1/3,所以,sina=2根号2/3,tana=-2根号2
sin(180°+α)-tanα/cos(180°-α)+tan(180°-α)
=-sina-(tana/-cosa)-tana
=-sina-tana+(tana/cosa)
=-(sina+tana)+(sina/cos^2 a)
=-(2根号2/3-2根号2)+(2根号2/3 /1/9)
=-(-4根号2/3)+6根号2
=4根号2/3+6根号2
=22/3根号2
不懂的欢迎追问,如有帮助请采纳,谢谢!
sin(180°+α)-tanα/cos(180°-α)+tan(180°-α)
=-sina-(tana/-cosa)-tana
=-sina-tana+(tana/cosa)
=-(sina+tana)+(sina/cos^2 a)
=-(2根号2/3-2根号2)+(2根号2/3 /1/9)
=-(-4根号2/3)+6根号2
=4根号2/3+6根号2
=22/3根号2
不懂的欢迎追问,如有帮助请采纳,谢谢!
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