y= sin3 x+sin6x x 1 1 cos(2x + 4 ) = cos x + 2 = C(x) (M2) Therefore, C(x) is periodic with period 2.

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1 . (a) y sin x+sin6x (A) (C) (b) Period (C) []. (a) y x 0 x O x Notes: Award for end points Award for a maximum of.5 Award for a local maximum of 0.5 Award for a minimum of 0.75 Award for the correct shape (b) C(x) cos x + cos x C(x + ) cos(x + ) + cos(x + 4) cos x + cosx C(x) (M) Therefore, C(x) is periodic with period. (c) C(x) is a maximum for x, 0, (A) Note: Do not penalize candidates who also write x,

2 (d) x 0. (using a graphic display calculator) (A) (e) (i) C( x) cos( x) + cos ( x) cos x + cosx C(x) for all x (AG) (ii) C(x 0 ) 0 so C( x 0 ) 0 [C(x) C( x)] C( x 0 ) C( x 0 ) 0 [C(x) is periodic, period ] Therefore, x x 0 (R). sin x tan x sin x cos x sin x 0 sin x( cos x ) 0 sin x 0, cos x x 0, x ± or ±.05 ( s. f.) (C) OR x 0, x ± (or ±.05 ( s. f.)) (G)(G)(G) (C) Note: Award (G) for x 0, ± 60. [] 4. (a) y arccos (. cos x) y arcsin (.4 sin x) (b) The solutions are x.6, y x 0.464, y.6 5. METHOD tan + tan tan + 0 tan 5 tan 0.8, , 69. (C6)

3 METHOD sin cos cos sin sin cos sin sin 0.9, 69. (C6) 6. METHOD cos l l cos (or l sin sin ) cos (or sin ) cos (or sin ) 0.65,.5 (accept 0.96, 0.804) (C)(C) METHOD cos ( cos ) cos θ 0.65,.5 (accept 0.96, 0.804) (C)(C) Notes: Do not penalize if the candidate has included extra solutions. Penalize [ mark] if candidates give answers in degrees, ie award for 5., 45 ; (A0) for one correct answer in degrees. 7. (a) t Either finding depths graphically, using sin or solving 6 h(t) 0 for t h (t) max (m), h (t) min 4 (m) N t (b) Attempting to solve 8 + 4sin 8 algebraically or graphically 6 t [0, 6] [, 8] {4} N 8. Since range goes from 4 to a Since curve is shifted right by, b 4 4 Since curve has been shifted in vertical by one unit down c

4 a b c NNN 4 9. tan θ 5 se cθ 0 0 Using l + tan θ sec θ, (sec θ l) 5 sec θ 0 0 sec θ 5 sec θ 0 Solving the equation eg ( sec θ + ) (sec θ 4) 0 sec θ or sec θ 4 θ in second quadrant sec θ is negative sec θ 0. EITHER tan tan OR (R), 4 4, or.8, 0.9 N4 4 8 N, or.8, x. 0 cm water depth corresponds to 6 sec 6 6 Rearranging to obtain an equation of the form sec equivalent x k or 6 ie making a trignometrical function the subject of the equation. M 4

5 x cos 6 x arccos 6 8 x arccos M Note: Do not penalize the omission of. Width of water surface is 7 8 arccos (cm) R N Note: Candidate who starts with 0 instead of 6 has the potential to gain the two M marks and the R mark.. (a) cosx + sinx R cos cosx + R sin sinx R cos, R sin R, Note: Award (A0) if degrees used instead of radians. (b) (i) Since f (x) cos x, f max when x ; f min (when x 0) Range is [, ] (ii) Inverse does not exist because f is not : (R) 5 Notes: Award (R) for a correct answer with a valid reason. Award (R) for a correct answer with an attempt at a valid reason, eg horizontal line test. Award (R0) for just saying inverse does not exist, without any reason. (c) f (x) cos x x OR f (x) 4 x 5

6 x 0.6 x (G) (d) I sec x dx 0. METHOD lnsec x tan x 0 ln ln ln ( + ). (AG) 5 b Note: Award zero marks for any work using GDC. ab a a b ()() cos ( cos ) 4sin sin. (C) METHOD 6

7 O a b A M ab B AB ab In OAM, AM OA sin. Therefore, a b sin. (C) [] 4. (a) f ( ) R cos cos + R sin sin R cos 4, R sin R 5, arctan f ( ) 5 cos ( 0.644) (C4) (b) f ( ) is maximum when ie radians (C) 5. (a) cos θ sin θ r cos (θ + α) where r and α arctan (or 0 ) 6 cos sin cos 6 (b) Since cos θ sin θ cos 6 range will be [, ]. (c) cos θ sin θ cos 6 cos 6 4 +, 6 7

8 θ 7, 6 Note: Answers must be multiples of. 5 [0] 6. sin 4 ( cos ) sin cos ( (cos sin )) cos ( cos4 ) cos ( (cos sin ) sin ( cos sin ) cos sin sin (sin ) sin sin sin sin sin cos sin cos tanθ (AG) [5] 7. (a) cos (A + B) cos A cos B sin A sin B, cos (A B) cos A cos B + sin A sin B Hence cos (A + B) + cos (A B) cos A cos B (AG) (b) (i) T (x) cos (arccos x) x (ii) T (x) cos ( arccos x) cos (arccos x) (A) x (AG) 5 (c) (i) T n+ (x) + T n (x) cos [(n + )arccos x] + cos [(n )arccos x] Using part (a) with A n arccos x, B arccos x T n+ (x) + T n (x) cos (n arccos x) cos (arccos x) x cos (arccos x) xt n (x) (AG) (ii) Let P n be the statement: T n (x) is a polynomial of degree, n T (x) x, a polynomial of degree one. So P is true. T (x) x, is a polynomial of degree two. So P is true. Assume that P k is true. From part (c)(i), T k+ (x) xt k (x) T k (x) Assume P k is true as well. T k (x) has degree k xt k (x) has degree (k + ) + 8

9 and as T k+ (x) has degree (k ) T k+ (x) has degree (k + ) By the principle of mathematical induction, P n is true for all positive integers n. (R) Notes: These arguments may be in a different order. There is a maximum of 6 marks in part (ii) for candidates who do not consider a two stage process. 8. a sin x cosx bsin x 0 sin x(a cos x b) 0 b cosx a cos x b a b b a b cos x a 4 a 4 a 9. (a) (i) (cos + i sin) cos + cos i sin + cos i sin + i sin Note: ( cos + cos sin i cos sin i sin ) ( cos cos sin + ( cos sin sin ) i) Award for each term in the expansion. (ii) (cos + i sin) cos + i sin equating real and imaginary parts (C6) cos cos cos sin cos cos ( cos ) 4 cos cos AG N0 and sin cos sin sin ( sin ) sin sin sin 4 sin AG N0 [9] (b) sin θ sin θ sin θ 4sin θ sin θ cosθ cos θ 4cos θ cos θ cos θ sin θ sin θ cos θ cos θ Using sin cos cos sin θ sin θ sin θ cosθ cos θ cos θ M tan AG 9

10 (c) METHOD sin θ cos θ sin θ 4 M 7 7 cosθ 4 M 0 7 tan θ N0 0 0 METHOD sin θ sin θ 4 M cos θ M 7 7 tan θ Aˆ Bˆ Cˆ M Aˆ Bˆ Cˆ N0 [] tan ( Â + Bˆ ) tan ( Ĉ ) M tan Aˆ tan Bˆ tan Cˆ tan Aˆ tan Bˆ tan Aˆ tan Bˆ tanĉ + tan Aˆ tan Bˆ tancˆ tan Aˆ tan Bˆ tan Cˆ tan Aˆ tan Bˆ tancˆ AG N0 0

11 . (a) Note: Award for shape. (i) x intercepts:.6,.6 (ii) Asymptotes: x,0, 4 4 (iii) Max. point ( 0.44,.56) Min. point (0.44,.56) (b) cosec x + tan x 0 M sin x 0 sin x cos x sin x cos x 0 sin x cos x cos x + sin x cos x 0 Note: of the above 4 marks may be implied. cos x + cos x cos x 0 cos x cos x cos x + 0 AG N0 (c) y cosec x + tan x dy cosec x cot x + sec x 0 for max/min M dx cos x 0 sin x sin x cos x cos x cos x + sin x 0 cos x ( cos x ) + ( cos x) 0 M

12 4cos 5 x + 4cos x cos x + cos x 0 (or equivalant) 4 cos 5 x 4 cos x + cos x + cos x 0 AG N0 f tan x M sin x (d) x tanx sin x f x tan x M sin x tanx sin x AG N0 sin x sin x Hence f x f x tanx tanx 0 [5]. METHOD AC 5 and AB (may be seen on diagram) cos α and 5 cos β and 4 sin α 5 sin β Note: Use of cos ( ) cos cos + sin sin If only the two cosines are correctly given award (A0). 4 (substituting) M N METHOD AC 5 and AB (may be seen on diagram) Use of cos ( + ) AC AB BC ACAB Use of cos ( + ) + cos ( ) cos cos cosα and 5 cos β

13 7 7 cos α N β (sin + i ( cos)) sin ( cos) + i sin ( cos) M Let be the required argument. tan sin θ cos θ θ sin θ cos sin θ cos θ cos θ cos θ cos θ sin θ cos θ cos θ cos θ M tan [7]

(e) (i) Prove that C(x) = C( x) for all x. (2)

(e) (i) Prove that C(x) = C( x) for all x. (2) Revision - chapters and 3 part two. (a) Sketch the graph of f (x) = sin 3x + sin 6x, 0 x. Write down the exact period of the function f. (Total 3 marks). (a) Sketch the graph of the function C ( x) cos