if sinθ=2/sqrt5 what is tanθ equal to?
回答 (4)
2016-12-05 9:59 pm
x² = 5 - 4 = 1
x = 1
sin Ө = 2 / √5
cos Ө = 1 / √5
tan Ө = 2
x = 1
sin Ө = 2 / √5
cos Ө = 1 / √5
tan Ө = 2
2016-12-05 8:48 pm
sinθ = 2/√5
cos²θ = 1 - sin²θ = ⅕
cosθ = √(⅕) = √5/5
tanθ = sinθ/cosθ = ⅖
cos²θ = 1 - sin²θ = ⅕
cosθ = √(⅕) = √5/5
tanθ = sinθ/cosθ = ⅖
2016-12-05 8:47 pm
sin(θ) = y/h = 2/sqrt(5)
x^2 + y^2 = h^2
x^2 + 2^2 = 5
x^2 = 1
x = +/- 1
cos(θ) = x/h = +/- 1/sqrt(5)
tan(θ) = sin(θ)/cos(θ) = (2/sqrt(5))/(+/- 1/sqrt(5)) = +/- 2
x^2 + y^2 = h^2
x^2 + 2^2 = 5
x^2 = 1
x = +/- 1
cos(θ) = x/h = +/- 1/sqrt(5)
tan(θ) = sin(θ)/cos(θ) = (2/sqrt(5))/(+/- 1/sqrt(5)) = +/- 2
2016-12-05 8:44 pm
rules
cos^2 x + sin^2 x = 1
tanx = sinx/cosx
cos^2 x + sin^2 x = 1
tanx = sinx/cosx
收錄日期: 2021-04-23 23:55:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161205123728AAova04