急！cosΘ＝2/7, sinΘ=?
回答 (5)
2009-06-15 11:56 am
✔ 最佳答案
有幾種方法。方法一，用畢氏定理：
對邊：√(7^2 - 2^2) = √45 = 3√5
∴ sinΘ = 3√5 / 7
方法二，使用公式：
cosΘ = 2/7
cos^2 Θ = (2/7)^2 = 4/49
sin^2 Θ = 1 - cos^2 Θ = 1 - 4/49 = 45/49
∴sinΘ = √(45/49) = √45/7 = 3√5/7
參考: me
2009-06-16 8:28 am
see foto
than u may understand more clear than just explain by words
http://i29.photobucket.com/albums/c255/sylki/9863f20e.jpg
than u may understand more clear than just explain by words
http://i29.photobucket.com/albums/c255/sylki/9863f20e.jpg
2009-06-15 9:35 pm
the opposite side = √(7^2-2^2) = √45
sinΘ =√45 / 7
= 0.95831(5 sig.fig.)
sinΘ =√45 / 7
= 0.95831(5 sig.fig.)
參考: me and my brain
2009-06-15 5:40 am
cosΘ＝2/7, sinΘ=?
Sol
(1)Θ在第一象限
cosΘ>0，sinΘ>0
sinΘ=sqrt(49-4)/7=sqrt(45)/7=3sqrt(5)/7
(2)Θ在第四象限
cosΘ>0，sinΘ<0
sinΘ=－sqrt(49-4)/7=－sqrt(45)/7=－3sqrt(5)/7
Sol
(1)Θ在第一象限
cosΘ>0，sinΘ>0
sinΘ=sqrt(49-4)/7=sqrt(45)/7=3sqrt(5)/7
(2)Θ在第四象限
cosΘ>0，sinΘ<0
sinΘ=－sqrt(49-4)/7=－sqrt(45)/7=－3sqrt(5)/7
2009-06-15 5:05 am
X= sqrt(7^2-2^2)
X=3 sqrt(5)
sinΘ= 3 sqrt(5) / 7
2009-06-14 21:06:46 補充：
sinΘ= 3√5 / 7
X=3 sqrt(5)
sinΘ= 3 sqrt(5) / 7
2009-06-14 21:06:46 補充：
sinΘ= 3√5 / 7
收錄日期: 2021-04-13 16:40:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090614000051KK01946