F.4 Amaths question (trigonometry)2

Let π < a < 3π/2. If sin a and sec a are the roots of the equation 2y² - ky +3 =0, find the value of the constant k. Leave your answer in surd form.

Product of roots = 3/2

sin a sec a = 3/2

sin a / cos a = 3/2

tan a = 3/2

Consider a triangle ABC with angle BAC = a, AB = 2, BC = 3,

the longest side AC

= √(3² + 2²)

= √13

So cos a = AB/AC = -2/√13 .... (1) 【因為 π < a < 3π/2 所以 cos a 是負數】

So sin a = BC/AC = -3/√13 .... (2)【因為 π < a < 3π/2 所以 sin a 是負數】

Sum of roots = -(-k)/2

sin a + sec a = k/2

sin a + 1/cos a = k/2

-3/√13 + 1/(-2/√13) = k/2

-3/√13 - √13/2 = k/2

6/√13 + √13 = -k

-k = 6/√13 + √13

-k = (6 + √13√13)/√13

-k = (6 + 13)/√13

-k = 19/√13

k = 19√13/13


2006-12-30 13:35:46 補充：
小小補充：這種問題 (通常答案是 surd form) 通常會有方法得到 sin a, cos a 或 tan a 的值。再通過公式或上面一般的三角形方法可得到所有 sin a, cos a 和 tan a 值。如果答案要是 surd form，切忌找出 a 的值（除非 a 的值是 30°, 60°, 等你有信心可以用公式轉回 surd form 的角度）。

2006-12-30 14:00:29 補充：
除了三角形方法，也給你一個公式例子。如上題 tan a = 3/2，sec²a=1 tan²asec²a=1 (3/2)²sec²a=13/4sec a= √13/2cos a = 1/sec acos a = 2/√13sin²a = 1 - cos²asin²a = 1 - (2/√13)²sin²a = 9/13sin²a = 3/√13一樣可以得到餘下的sin a 和 cos a 值。所以只是視乎你用三角形方法還是公式方法。

2y^2 - ky + 3 =0

product of roots = (sin a) (sec a)
3/2 = tan a
a = (56.31° + 180°)
a = 236.31°

sum of roots = sin a + sec a
k/2 = -0.83205 - 1.80278
k = -5.2697

[-19(√13)] / 13 = -5.2697