Page 52 - ACE YR IGCSE A TOP APPR' TO ADD MATH

(b) sin x = cos 2x + 1 (d) sec 1 3a – 18° = –3
2
2 solutions 2
[2] 3a 1
2
cos 1 – 18° = –
12. (a) Amplitude = 5 2 3
360° π 0° < a < 360°
Period = = 90° =
4 2 3a
[3] −18° < 2 − 18° < 522°
2
(b) Maximum point 1 π , 7 3a − 18° = 109.47°, 250.53°, 469.47°
2
2
a = 84.98°, 179.02°, 324.98°
2
2
Minimum point 1 π , –3 and 1 3π , –3 [2] [4]
4
4
2
2
(b) Maximum point 1 π , 8 and 1 5π , 8 13. (a) Amplitude = 3 16. When x = 2
2(2) + (2) − 13(2) + 6 = 0
2
3
6
6
3
2
2
2
Minimum point 1 π , 2 Period = 120° [3] (2x + x − 13x + 6) ÷ (x − 2) = 2x + 5x − 3
2x + x − 13x + 6 = (x − 2)(2x − 1)(x + 3)
2
3
2
(b) Maximum point (30°, 8) and (150°, 8) 2 tan y + tan y – 13 = – 6
2
Minimum point (90°, 2) tan y
3
2
[2] 2 tan y + tan y − 13 tan y + 6 = 0
3
14. y = 1 + 3 sin x (tan y − 2)(tan y + 3)(2 tan y − 1) = 0
1
2
tan y = 2 or tan y = or tan y = −3
3
a = 3, b = , c = 1 2
2 [3] y = 26.57°, 63.43°, 108.43°, 206.57°, 243.43°, 288.43°
[6]
15. (a) −4 sin x + 8 sin x − 5 + 5 sin x = 0 17. (a) 5 sin q − 4 cos q = 5 cos q + 4 sin q
2
2
sin x + 8 sin x − 5 = 0 sin q = 9 cos q
2
2
sin x = –8 ± √8 – 4(1)(–5) sin q = 9
2(1) cos q
= –4 + √21 or –4 – √21 (rejected) tan q = 9
q = 263.66°
sin x = –4 + √21 [3]
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x = 35.63°, 144.37°
(b) (5 sin q − 4 cos q) + (5 cos q + 4 sin q)
2
2
[4] = 25 sin q − 40 sin q cos q + 16 cos q +
2
2
2
2
(b) sin 2x = – 2 25 cos q + 40 cos q sin q + 16 sin q
3 = 41 sin q + 41 cos q
2
2
0° < x < 360° = 41
0° < 2x < 720° [3]
2x = 221.81°, 318.19°, 581.81°, 678.19° 18. (a) 1 – 3 sin x = – cos x
x = 110.9°, 159.1°, 290.9°, 339.1° sin x sin x
2
[4] 1 – 3 sin x = – cos x
(c) sec y + sec y tan y = 5 cosec y sin x sin x
2
1 + tan y = 5 1 − 3(1 − cos x) = −cos x
2
cos y cos y sin y 1 − 3 + 3 cos x + cos x = 0

3 cos x + cos x − 2 = 0
2
1 + tan y = 5 (3 cos x − 2)(cos x + 1) = 0
cos y sin y 2
5 cos y cos x = or cos x = −1
1 + tan y = 3
sin y x = 48.19°, 311.81°, 180°
1 + tan y = 5 [4]
tan y (b) 2(1 + cot q) = 14 − 2 cot q
2
2
tan y + tan y – 5 = 0 2 + 2 cot q − 14 + 2 cot q = 0
2
2 cot q − 2 cot q − 12 = 0

2
tan y = – 1 + √21 or tan y = –1 – √21 (cot q − 3)(cot q + 2) = 0
2 2 1 1
y = 60.83°, 109.71°, 240.83°, 289.71° tan q = 3 or tan q = –2
[4] q = 18.43°, 153.43°, 198.43°, 333.43°
[4]
Cambridge IGCSE
TM
192 Ace Your Additional Mathematics
Answers Add Math.indd 192 14/03/2022 12:29 PM