Let (-7,4) be a point on the terminal side of θ. Find the exact values of cosθ cscθ and tanθ?
回答 (2)
2016-11-08 6:00 pm
a = -7, o = 4; given h² = o² + a² then h = √((-7)² + 4²) = √65
sin(θ) = o/h = 4/√65 aka 4√(65)/65
cos(θ) = a/h = -7/√65 aka -7√(65)/65
tan(θ) = o/a = 4/(-7) = -4/7
csc(θ) = h/o = √(65)/4
sec(θ) = h/a = √(65)/(-7) = -√(65)/7
cot(θ) = a/o = -7/4
sin(θ) = o/h = 4/√65 aka 4√(65)/65
cos(θ) = a/h = -7/√65 aka -7√(65)/65
tan(θ) = o/a = 4/(-7) = -4/7
csc(θ) = h/o = √(65)/4
sec(θ) = h/a = √(65)/(-7) = -√(65)/7
cot(θ) = a/o = -7/4
2016-11-08 6:33 pm
(-7,4) is a point on the terminal side of t. Therefore t is in Q2 with cos(t) =
-7/(49+16)^(1/2) = -(7/65)rt65. sin(t) = +[1 - cos^2(t)]^(1/2) =
[1 - 49/65]^(1/2) = (4/65)rt65. csc(t) =1/sin(t) =(1/4)rt65. tan(t) =sin(t)/cos(t)
= - (4/7).
-7/(49+16)^(1/2) = -(7/65)rt65. sin(t) = +[1 - cos^2(t)]^(1/2) =
[1 - 49/65]^(1/2) = (4/65)rt65. csc(t) =1/sin(t) =(1/4)rt65. tan(t) =sin(t)/cos(t)
= - (4/7).
收錄日期: 2021-04-21 23:59:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161108094048AAmFmF5