A Maths Trigo

2011-02-09 4:31 am
1. Let cos(a+b) = p , sin(a+c) = q.
a) Express cos(a) and sin(a) n terms of b,c,p,q .
b) Hence show that p^2 + q^2 + 2pq sin(b-c) = cos^2(b-c).

2.If the equation 4(cosA - sinA) + ksinA=5 has no solution, find the range of values of k.

回答 (1)

2011-02-09 5:07 pm
✔ 最佳答案
1a) Expanding:

cos a cos b - sin a sin b = p ... (1)

sin a cos c + cos a sin c = q ... (2)

(1) x sin c: cos a cos b sin c - sin a sin b sin c = p sin c ... (3)

(2) x cos b: sin a cos b cos c + cos a cos b sin c = q cos b ... (4)

(4) - (3):

sin a cos b cos c + sin a sin b sin c = q cos b - p sin c

sin a (cos b cos c + sin b sin c) = q cos b - p sin c

sin a = (q cos b - p sin c)/cos (b - c)

(1) x cos c: cos a cos b cos c - sin a sin b cos c = p cos c ... (5)

(2) x sin b: sin a sin b cos c + cos a sin b sin c = q sin b ... (6)

(5) + (6):

cos a cos b cos c + cos a sin b sin c = p cos c + q sin b

cos a (cos b cos c + sin b sin c) = p cos c + q sin b

cos a = (p cos c + q sin b)/cos (b - c)

b) sin2 a + cos2 a = 1

(p sin c - q cos b)2/cos2 (b - c) + (p cos c + q sin b)2/cos2 (b - c) = 1

(p sin c - q cos b)2 + (p cos c + q sin b)2 = cos2 (b - c)

p2 sin2 c + q2 cos2 b - 2pq sin c cos b + p2 cos2 c + q2 sin2 b + 2pq cos c sin b = cos2 (b - c)

p2 + q2 + 2pq (cos c sin b - sin c cos b) = cos2 (b - c)

p2 + q2 + 2pq sin (b - c) = cos2 (b - c)

2) (k - 4) sin A + 4 cos A = 5

R sin (A + θ) = 5 where R = √[(k - 4)2 + 42] = √(k2 - 8k + 32)

So for the equation to have no soln, √(k2 - 8k + 32) > 5, i.e.

k2 - 8k + 32 > 25

k2 - 8k + 7 > 0

(k - 7)(k - 1) > 0

k > 7 or k < 1
參考: 原創答案


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