Trig Equation with Reciprocals?
回答 (3)
2016-07-29 9:25 am
When θ is in the fourth quadrant, i.e. 270° < θ < 360° :
sinθ < 0
secθ = √5
1/cosθ = √5
cosθ = 1/√5
sin²θ + cos²θ = 1
sin²θ + (1/√5)² = 1
sin²θ + (1/5) = 1
sin²θ = 4/5
As sinθ < 0, then sinθ = -√(4/5)
sinθ = -√(20/25)
sinθ = -2(√5)/5
sinθ < 0
secθ = √5
1/cosθ = √5
cosθ = 1/√5
sin²θ + cos²θ = 1
sin²θ + (1/√5)² = 1
sin²θ + (1/5) = 1
sin²θ = 4/5
As sinθ < 0, then sinθ = -√(4/5)
sinθ = -√(20/25)
sinθ = -2(√5)/5
2016-07-29 4:18 pm
secθ = √5
1/cosθ = √5
cosθ = 1/√5
sin²θ = 1 - cos²θ = ⅘
sinθ = -√(⅘) = -2/√5
1/cosθ = √5
cosθ = 1/√5
sin²θ = 1 - cos²θ = ⅘
sinθ = -√(⅘) = -2/√5
2016-07-29 9:29 am
OK, since sec is positive in quadrant IV, well take the positive value. How does that sound?
secθ = √5
1/cosθ = √5
cosθ = 1/√5
Now use the Pythagorean identity :
sin^2 (θ) + cos^2(θ) = 1
sinθ = √(1 - cos^2 (θ))
sinθ = √(1 - 1/5)
sinθ = √(4/5)
sinθ = 2/√5
Hope this helps :)
secθ = √5
1/cosθ = √5
cosθ = 1/√5
Now use the Pythagorean identity :
sin^2 (θ) + cos^2(θ) = 1
sinθ = √(1 - cos^2 (θ))
sinθ = √(1 - 1/5)
sinθ = √(4/5)
sinθ = 2/√5
Hope this helps :)
收錄日期: 2021-04-18 15:21:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160729003218AAEfG3T