sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=2008,则tan(x+225)的值为多少是sin加sin再除以cos=sin
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![sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=2008,则tan(x+225)的值为多少是sin加sin再除以cos=sin](/uploads/image/z/5191100-44-0.jpg?t=sin%EF%BC%8890-x%EF%BC%89%2Bsin%EF%BC%88180-x%EF%BC%89%2Fcos%28-x%29%2Bsin%28-x%29%3D2008%2C%E5%88%99tan%28x%2B225%29%E7%9A%84%E5%80%BC%E4%B8%BA%E5%A4%9A%E5%B0%91%E6%98%AFsin%E5%8A%A0sin%E5%86%8D%E9%99%A4%E4%BB%A5cos%3Dsin)
sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=2008,则tan(x+225)的值为多少是sin加sin再除以cos=sin
sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=2008,则tan(x+225)的值为多少
是sin加sin再除以cos=sin
sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=2008,则tan(x+225)的值为多少是sin加sin再除以cos=sin
sin(90-x)=cosx
sin(180-x)=sinx
cos(-x)=cos x
sin(-x)=-sin x
∴化为(cosx+sinx)/(cosx-sinx) 分式上下同除cosx得 (1+tanx)/(1-tanx)=2008 得 tanx=2007/2009
∵ tanx周期为π,∴tan(x+225)=tan(x+45)=(tanx+tan45)/1-tanx.tan45=2008
sin(90-x)+sin(180-x)/cos(-x)+sin(-x)=(cosx+sinx)/(cosx-sinx)
分子分母同时除以cosx(由题可知cosx≠0)
得(1+tanx)/(1-tanx)=(tan45°+tanx)/(1-tanx*tan45°)=tan(x+45°)=2008
tan(x+225°)=tan(180°+x+45°)=tan(x+45°)=2008
sin(90-x)=cosx
sin(180-x)=sinx
cos(-x)=cos x
sin(-x)=-sin x
所以原式=(cosx+sinx)/(cosx-sinx) 分式上下同除cosx得 (1+tanx)/(1-tanx)=2008 解得 tanx=2007/2009
因为 tanx周期为π,所以...
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sin(90-x)=cosx
sin(180-x)=sinx
cos(-x)=cos x
sin(-x)=-sin x
所以原式=(cosx+sinx)/(cosx-sinx) 分式上下同除cosx得 (1+tanx)/(1-tanx)=2008 解得 tanx=2007/2009
因为 tanx周期为π,所以 tan(x+225)=tan(x+45)=(tanx+tan45)/1-tanx.tan45=2008
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