急！cosΘ＝2/7, sinΘ=?

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

see foto
than u may understand more clear than just explain by words

http://i29.photobucket.com/albums/c255/sylki/9863f20e.jpg

the opposite side = √(7^2-2^2) = √45


sinΘ =√45 / 7
= 0.95831(5 sig.fig.)

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

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