Let θ be an angle in quadrant 2 such that sinθ=5 over 6 Find the exact values of secθ and cotθ?
回答 (3)
2016-11-08 5:31 pm
sinθ = 5/6
cos²θ = 1-sin²θ = 11/36
Since θ is in quadrant II, cosθ = -√cos²θ
cosθ = -√(11/36) = -√11/6
secθ = 1/cosθ = -6/√11
cotθ = cosθ/sinθ = -√11/5
cos²θ = 1-sin²θ = 11/36
Since θ is in quadrant II, cosθ = -√cos²θ
cosθ = -√(11/36) = -√11/6
secθ = 1/cosθ = -6/√11
cotθ = cosθ/sinθ = -√11/5
2016-11-08 5:38 pm
If θ is in second quadrant
Then x = -√ 11, y = 5 and r = 6
secθ = 1/cosθ = 1/(-√11/6)
= -6/√11
cotθ = 1/tan(θ)
= 1/(5/-√11)
= -√11/5
Then x = -√ 11, y = 5 and r = 6
secθ = 1/cosθ = 1/(-√11/6)
= -6/√11
cotθ = 1/tan(θ)
= 1/(5/-√11)
= -√11/5
2016-11-08 5:33 pm
sin(θ) = y/h = 5/6
cos(θ) = x/h = x/6
x^2 + y^2 = h^2
x^2 + 5^2 = 6^2
x = 36 - 25
x = +/-sqrt(11)
In quad II, x is negative
x = -sqrt(11)
cos(θ) = -sqrt(11)/6
sec(θ) = 1/cos(θ) = -6/sqrt(11)
cot(θ) = cos(θ)/sin(θ) = (-sqrt(11)/6)/(5/6) = -sqrt(11)/5
cos(θ) = x/h = x/6
x^2 + y^2 = h^2
x^2 + 5^2 = 6^2
x = 36 - 25
x = +/-sqrt(11)
In quad II, x is negative
x = -sqrt(11)
cos(θ) = -sqrt(11)/6
sec(θ) = 1/cos(θ) = -6/sqrt(11)
cot(θ) = cos(θ)/sin(θ) = (-sqrt(11)/6)/(5/6) = -sqrt(11)/5
收錄日期: 2021-04-21 23:58:55
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