急! F4 Basic Trigonometry ex5

1.
180° ≤ θ ≤270°
360° ≤ 2θ ≤ 540°
(360° -90°) ≤ (2θ -90°) ≤ (540° - 90°)
270° ≤ (2θ -90°) ≤ 450°
0 ≤ cos(2θ - 90°) ≤ 1
(2 + 0) ≤ [2 + cos(2θ - 90°)] ≤ (2 + 1)
2 ≤ y ≤ 3

Maximum value of y = 3
Minimum value of y = 2


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2.
90° ≤ θ ≤ 360°
(90°+ 180°) ≤ (θ +180°) ≤ (360° +180°)
(90°+ 180°)/2 ≤ (θ +180°)/2 ≤ (360° +180°)/2
135° ≤ (θ + 180°)/2 ≤ 270°
-1 ≤ sin[(θ + 180°)/2] ≤ (√2)/2
4*(-1) ≤ 4sin[(θ + 180°)/2] ≤ 4*(√2)/2
-4 ≤ y ≤ 2√2

Maximum value of y = 2√2
Minimum value of y = -4

2012-10-21 01:36:55 補充：
sin 135° = sin (180 - 45)° = sin 45° = (√2)/2
sin 180° = 0
sin 270° = sin (180 + 90)° = -sin 90° = -1

When the angle changes from 135° to 270°, the sine value of the angle decreases from (√2)/2 to 1.

1.
180° ≤ θ ≤270°
360° ≤ 2θ ≤ 540°
270° ≤ 2θ -90° ≤ 450°
0 ≤ cos(2θ - 90°) ≤ 1
2 ≤ 2 + cos(2θ - 90°) ≤ 3
2 ≤ y ≤ 3

2.
90° ≤ θ ≤ 360°
270° ≤ θ +180° ≤ 540°
135° ≤ (θ + 180°)/2 ≤ 270°
-1 ≤ sin[(θ + 180°)/2] ≤ (√2)/2
-4 ≤ 4sin[(θ + 180°)/2] ≤ 2√2
-4 ≤ y ≤ 2√2

the key step in question 2
我地唔係話一定要對位，135出-1咁，而係考慮整個range入面，可能得出的最大/最小值
例：
-1<x<0
0<x^2<1
即係唔一定對位
-2<x<1
0≤x^2<4

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