IB SL: Trig Function Practice Answers

Transcription

1 IB SL: Trig Function Practice Answers. π From sketch of graph y = 4 sin x (M) or by observing sin. k > 4, k < 4 (A)(A)(C)(C) METHOD cos x = sin x cos x (M) cos x sin x cos x = 0 cos x(cos x sin x) = 0 (M) cos x = 0, (cos x sin x) = 0 (A)(A) π π x =, x = 4 (A)(A) (C6) METHOD Graphical solutions THEN EITHER for both graphs y = cos x, y = sin x, for the graph of y = cos x sin x. (M) (M) Points representing the solutions clearly indicated (A).57, (A) π π x =, x = (A)(A) (C6) 4 Notes: If no working shown, award (C4) for one correct answer. Award (C)(C) for each correct decimal answer.57, Award (C)(C) for each correct degree answer 90, 45. Penalize a total of [ mark] for any additional answers.

2 . (a) 0º Acute angle 0 (M) Note: Award the (M) for 0 and/or quadrant diagram/graph seen. nd quadrant since sine positive and cosine negative = 50 (A) (C) (b) tan 50 = tan 0 or tan 50 = (M) tan 50 = (A) (C) 4. (a) sin x = ( cos x) = cos x = l + cos x (M) => cos x + cos x l = 0 (A) (C) Note: Award the first (M) for replacing sin x by cos x. (b) cos x + cos x = ( cos x )(cos x +) (A) (C) (c) cos x = or cos x = l => x = 60, 80 or 00 (A)(A)(A) (C) Note: Award (A)(A)(A0) if the correct answers are given in 5 radians (ie,,, or.05,.4, 5.4)

3 5. AB = r r = r (M)(A) = (A) = 8 cm (A) (5.4) =.6 4 =.7 (=.48 radians) (M) AB = r (A) 4 = (M) = 8 cm (A) (C4) 6. (a) x is an acute angle => cos x is positive. (M) cos x + sin x = => cos x = sin (M) => cos x = (A) x 8 = (= ) (A) (C4) 9 (b) cos x = sin x = (M) 7 = 9 (A) (C) Notes: (a) Award (M)(M0)(A)(A0) for cos sin = (b) Award (M)(A0) for cos sin =

4 7. (a) area of sector ΑΒDC = π() = π 4 (A) area of segment BDCP = π area of ABC (M) = π (A) (C) (b) BP = (A) area of semicircle of radius BP = π( ) = π (A) area of shaded region = π (π ) = (A) (C) 8. (a) ( sin x) + sin x = (A) 6 sin x sin x = 0 (p = 6, q =, r = ) (A) (C) (b) ( sin x )( sin x + ) (A)(A) (C) (c) 4 solutions (A) (C) 9. cos x = 5 sin x sin x cos x 5 (M) tan x = 0.6 (A) x = or x = (to the nearest degree) (A)(A)(C)(C) Note: Deduct [ mark] if there are more than two answers. 0. Perimeter = 5(π ) + 0 (M)(A)(A) Note: Award (M) for working in radians; (A) for π ; (A) for +0. = (0π + 5) cm (= 6.4, to sf) (A) (C4) 4

5 . (a) p = 0 A (b) METHOD Period = q (M) = (A) q = 4 A 4 METHOD Horizontal stretch of scale factor = q (M) scale factor = 4 (A) q = 4 A 4. METHOD The value of cosine varies between and +. Therefore: t = 0 a + b = 4. t = 6 a b = 0. a = 4.6 a =. (A) (C) b = 4.0 b = (A) (C) π() Period = hours k = π (M) k = (A) (C) METHOD y t (h) From consideration of graph: Midpoint = a =. (A) (C) Amplitude = b = (A) (C) Period = π π = (M) k k = (A) (C) 5

6 . (a) cos x + sin x = ( sin x) + sin x = sin x + sin x (A) (b) cos x + sin x = sin x + sin x = sin x sin x = 0 sin x( sin x) = 0 sin x = 0 or sin x = sin x = 0 x = 0 or (0 or 80 ) Note: Award (A) for both answers. (M) (A) π 5π sin x = x = or (0 or 50 ) (A) 6 6 Note: Award (A) for both answers. 4. OTˆ A = 90 (A) AT = 6 = 6 π TÔA = 60 = (A) Area = area of triangle area of sector π = (M) =. cm (or 8 6) (A) (C4) TÔA = 60 (A) Area of = 6 sin 60 (A) π Area of sector = 6 6 (A) Shaded area = 8 6 =. cm ( sf) (A) (C4) 6

7 5. (a) (i) sin =.4 (A) (ii) At 00, t = (A) sin 0.5 = 6.48 (A) (N) Note: Award (A0)(A) if candidates use t = 00 leading to y =.6. No other ft allowed. (b) (i) 4 metres (A) (ii) t t 4 = sin sin = (M) t = π (.4) (correct answer only) (A) (N) (c) (i) 4 (A) (ii) t sin = 7 (M) t sin = 0.75 (A) t = 7.98 (A) (N) (iii) depth < 7 from 8 = hours (M) from 00 0 = hours (M) therefore, total = 6 hours (A) (N) 7 [] 7

8 6. (a) (i) 7 A N (ii) A N (iii) 0 A N (b) (i) evidence of appropriate approach M eg A = 8 AG N0 (ii) C = 0 A N (iii) METHOD period = evidence of using B period = (accept 60) eg = B 6 METHOD evidence of substituting eg 0 = 8 cos B + 0 simplifying eg cos B = 0 B 6 8 A B accept accept 0.54or0 B 0.54or0 (A) (M) A (M) (A) A N N (c) correct answers AA eg t =.5, t = 0.5, between 0: and 0:9 (accept 0:0) N [] 8

9 Trig Quest Answers sin B sin 48. Using sine rule: (M)(A) sin B = sin 48 = (M) 7 B = arcsin (0.508) =.06 (M)(A) = (nearest degree) (A) (C6) Note: Award a maximum of [5 marks] if candidates give the answer in radians (0.560).. Note: Award (M) for identifying the largest angle cos = 4 5 (M) = 5 (A) = 0.5 (A) Find other angles first = 44.4 = 4.0 (M) = 0.6 (A)(A) (C4) Note: Award (C) if not given to the correct accuracy. PQ. (a) = tan 6 40 PQ 9. m ( sf) (A) (C) (b) 40m 0 B Q AQˆ B 70 A = 80 (A) AB 40 sin 80 sin 70 AB = 4 9. m ( sf) Note: Award (M) for correctly substituting. (M) (A) (C) 9

10 4. A d B P (M)(A).5 0 = 50 (M)(A).5 = 80 (A) d = cos 70 d = 78.5 km (M)(A) (A) (C6) 5. C km A 48km B cos CÂB (M)(A) (48)() CÂB = arccos(0.065) (A) 86 (A) sin (AĈB) sin (a) (M) 0 7 0sin 50 sin (AĈB) = AĈB > 90 AĈB = = 5.7 AĈB = 6 ( sf) (A) (C) (b) In Triangle, AĈB = 64. BÂC = 80 ( ) = 65.7 (A) Area = (0)(7) sin 65.7 = 55 (cm ) ( sf) (A) (C) 0

11 7. (a) The smallest angle is opposite the smallest side cos θ = (M) = = Therefore, θ = 8. (A) (C) (b) Area = 8 7 sin 8. (M) = 7. cm (A) (C) (a) (M) sin A sin 45 sin A = 6 (A) 7 6 = (AG) 7 (b) A D h B C (i) BDˆ C + BÂC = 80 (A) 6 (ii) sin A = 7 => A = 59.0 or ( sf) (A)(A) => BĈD = 80 ( + 45 ) = 4.0 ( sf) (A) (iii) 7 BD sin 4 sin 45 (M) =>BD =.69 (A) 6

12 (c) BD h Area BDC Area BAC BA h BD (M)(A) = BA (AG) BD 6sin 45 Area ΔBCD Area ΔBAC BA 6sin 45 (M)(A) BD = (AG) BA [0] 9. (a) Sine rule PR 9 sin5 sin0 (M)(A) 9sin 5 PR = sin0 = 5.96 km (A) (b) EITHER Sine rule to find PQ 9sin 5 PQ = sin0 (M)(A) = 4.9 km (A) Cosine rule: PQ = ()(5.96)(9) cos 5 = 9.9 PQ = 4.9 km Time for Tom = (M)(A) (A) (A) 5.96 Time for Alan = a (A) Then = 8 a (M) a = 0.9 (A) 7

13 (c) RS = 4QS (A) 4QS = QS QS cos 5 (M)(A) QS QS 8 = 0 (or x x 8 = 0) (A) QS = 8.0 or QS =.9 (G) therefore QS =.9 (A) QS QS sinsrˆ Q sin5 (M) SRˆ Q sin sin 5 (A) SRˆ Q = 6.7 (A) QŜR Therefore, = 80 ( ) = 8. (A) 9 QS SR sin8. sin6.7 sin 5 (M) 9sin6.7 9sin 5 QS = sin8. sin8. =.9 (A) 6