f.4 a.maths (20points!!)

hoi kei

2009-03-03 2:36 am

It is given that 9cosθ- 40sinθ= rcos(θ+β),where r>0 andβis
an acute angle. find:

(a)Let x= 1÷ ( ｜9cosθ- 40sinθ｜ + 19), find the range of values
of x. ( " ｜"呢個係 absolute value)

(b)Let y = (9cosθ- 40sinθ)÷(9cosθ- 40sinθ+50),find the range of values of y.

THX~!請列steps~

回答 (1)

六呎將軍

2009-03-03 5:30 am

✔ 最佳答案

(a) 9 cosθ- 40 sinθ= √(92 + 402) [cos θ/√(92 + 402) - 40 sin θ/√(92 + 402)]
= 41 [(cos θ)/41 - (40 sin θ)/41]
= 41 cos (θ + 88.6)
So, 0 <= |9 cosθ- 40 sinθ| <= 41
19 <= |9 cosθ- 40 sinθ| + 19 <= 60
1/60 <= 1/(|9 cosθ- 40 sinθ| + 19) <= 1/41
(b) y = (9cosθ- 40sinθ)/(9cosθ- 40sinθ+ 50)
= 1 - 50/(9cosθ- 40sinθ+ 50)
= 1 - 50/[41 cos (θ + 88.6) + 50]
-41 <= 41 cos (θ + 88.6) <= 41
9 <= 41 cos (θ + 88.6) + 50 <= 91
1/91 <= 1/[41 cos (θ + 88.6) + 50] <= 1/9
50/91 <= 50/[41 cos (θ + 88.6) + 50] <= 50/9
-50/91 >= -50/[41 cos (θ + 88.6) + 50] >= -50/9
41/91 >= 1 - 50/[41 cos (θ + 88.6) + 50] >= -41/9

參考: Myself

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