Given that tanθ=-15/112,find sinθ and cosθ
θ lies on quadrant II and IV
Let P(x,y) be a point on the terminal side of θ and OP =r
Let y=-15 and x=112
r^2=12769
r=113 or -113
(why -113 does not need to rejected?)
r must be positive?
More about Trigonometry
2012-04-18 1:01 am
回答 (2)
2012-04-18 2:30 am
✔ 最佳答案
tanθ=x/y=-15/112--> r=√(x^2+y^2)=113sinθ=112/113, cosθ=-15/113~第二象限
sinθ=-112/113, cosθ=15/113~第三象限
r 必須是正的
2012-04-17 20:44:14 補充：
有誤 修正如下
tanθ=x/y=-15/112--> r=√(x^2+y^2)=113
sinθ=15/113, cosθ=-112/113~第二象限
sinθ=-15/113, cosθ=112/113~第三象限
r 必須是正的
2012-04-18 1:43 am
問題錯?
radius只會是正數
radius只會是正數
收錄日期: 2021-04-20 12:37:31
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