M2 trigonometry =_=

2010-08-30 10:47 pm

只須做(b)題

(a) Let t = tan( θ /2). show that

(1)Tanθ= 2t / (1 - t^2) (2)Sinθ= 2t / (1 - t^2) (3)Cosθ= (1 - t^2) / (1 + t^2) (b) Given the equation sin^2 x + sin2x = m + 2cos^2 x, where m is a real number,

(1)Express the equation in terms of sin2x and cos2x.
(2)Let t = tanx. Show that (m-1)(t^2) – 2t + (m+2) = 0.
(3)If the equation has real solutions, find the possible range of values of m.

回答 (1)

六呎將軍

2010-08-30 11:16 pm

✔ 最佳答案

b1) Applying the formulae:

sin2 x = (1 - cos 2x)/2 and cosx = (1 + cos 2x)/2

We have:

(1 - cos 2x)/2 + sin 2x = m + 2[(1 + cos 2x)/2]

1 - cos 2x + 2 sin 2x = 2m + 2 + 2 cos 2x

3 cos 2x - 2 sin 2x + 2m + 1 = 0

b2) Applying half-angle formula:

cos 2x = (1 - t2)/(1 + t2) and sin 2x = 2t/(1 + t2)

So the equation becomes:

3(1 - t2)/(1 + t2) - 4t/(1 + t2) + 2m + 1 = 0

3(1 - t2) - 4t + 2m(1 + t2) + 1 + t2 = 0

3 - 3t2 - 4t + 2m + 2mt2 + 1 + t2 = 0

(2m - 2)t2 - 4t + (2m + 4) = 0

(m - 1)t2 - 2t + (m + 2) = 0

b3) Since the equation has real solutions, the discriminant should be non-negative, i.e.

(-2)2 - 4(m - 1)(m + 2) >= 0

1 - (m2 + m - 2) >= 0

m2 + m - 3 <= 0

(m + 1/2)2 - 13/3 < = 0

(m + 1/2)2 <= 13/3

-√(13/3) <= m + 1/2 <= √(13/3)

-1/2 - √(13/3) <= m <= -1/2 + √(13/3)

參考: Myself

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