已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,则tan(A-圆周率/4)=?
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![已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,则tan(A-圆周率/4)=?](/uploads/image/z/5192411-59-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28cosA%2C-2%29%2Cb%3D%28sinA%2C1%29%E4%B8%94%E5%90%91%E9%87%8Fa.b%E5%B9%B3%E8%A1%8C%2C%E5%88%99tan%28A-%E5%9C%86%E5%91%A8%E7%8E%87%2F4%29%3D%3F)
已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,则tan(A-圆周率/4)=?
已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,则tan(A-圆周率/4)=?
已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,则tan(A-圆周率/4)=?
cosa/-2=sina/1
-0.5=sina/cosa
tana=-0.5
tan(a-π/4)=(tana-tanπ/4)/(1+tana*tanπ/4)=-1.5/0.5=-3
向量a=(cosA,-2),b=(sinA,1)且向量a.b平行,
所以
sinA/cosA=-1/2=tanA
tan(A-圆周率/4)=(tanA-tanπ/4)/(1+tanAtanπ/4)
=(-1/2-1)/(1-1/2)
=-3
因a//b,则:cosA/sinA=-1/2,即:tanA=-1/2,tan(A-π/4)=[tanA-tan(π/4)]/[1+tanAtan(π/4)]=[tanA-1]/[1+tanA]=-3
向量平行,对应坐标成比例,因为有数字,排除0向量可能。
cosA*1=sinA*-2
tanA=-1/2
tan(A-π/4)=(-1/2-1)/1+(-1/2)*1=-3
向量a.b平行,则cosA/-2=sinA/1得:tanA= -1/2
tan(A-π/4)=(tanA-tanπ/4)/(1+tanA*tanπ/4)=(-3/2)/(1/2)= -3