F.4---m2 trigonometry
2009-12-06 8:09 am
Suppose that α is a root of the following simultaneous equations in x.
{x^2 - 4xsinθ -2 =0
{x^2 - 4xcosθ +2 =0
(a) Show that α = 1 / (cosθ-sinθ)
(b) Hence, show that sin^2 θ = 1/4
回答 (1)
2009-12-06 8:18 am
✔ 最佳答案
(a) {α^2 - 4αsinθ - 2 = 0 ... (1){α^2 - 4αcosθ + 2 = 0 ... (2)
(2) - (1) => 4αsinθ - 4αcosθ + 4 = 0
4 = 4α(cosθ - sinθ)
α = 1/(cosθ - sinθ)
(b) Sub α = 1/(cosθ - sinθ) into (1),
[1/(cosθ - sinθ)]^2 - 4sinθ/(cosθ - sinθ) - 2 = 0
1 - 4sinθ(cosθ - sinθ) - 2(cosθ - sinθ)^2 = 0
1 - 4sinθcosθ + 4sin^2θ - 2cos^2θ + 4sinθcosθ - 2sin^2θ = 0
1 + 4sin^2θ - 2 = 0
4sin^2θ = 1
sin^2θ = 1/4
收錄日期: 2021-04-13 16:58:10
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