1. (a) = tanθ (as required) AG A1 cso 2. Note

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1 sinθ cosθ. (a) + cos θ sinθ cosθ cosθ cosθ Note tanθ (as required) AG A cso : Uses both a correct identity for sin θ and a correct identityfor cosθ. Also allow a candidate writing +cosθ cos θ on the denominator. Also note that angles must be consistent in when candidates apply these identities. A: Correct proof. No errors seen. (b) tanθ tanθ θ awrt 6.6 A θ awrt 5. Note A st for either tanθ or tanθ, seen or implied. A: awrt 6.6 Aft: awrt 5. or θ 80 + θ Special Case: For candidate solving, tanθ k, where, k, to give θ and θ 80 + θ, then award M0A0B in part (b). Special Case: Note that those candidates who writes tanθ, and gives ONLY two answers of 5 and 5 that are inside the range will be awarded SC M0A0B. [5]. (a) R 6.5 or.5 B tanα α awrt 0.65 A Note.5 B: R.5 or R 6.5. For R ±.5, award B0.. 5 : tan α ± or tan α ±. 5 A: α awrt 0.65

2 (b) (i) Max Value.5 Bft (ii) sin ( 0.65) Note π θ or θ theirα ; θ awrt. ; Aft B :.5 or follow through the value of R in part (a). : For sin (θ their α) π A : awrt. or + their α rounding correctly to sf. (c) H Max 8.5 (m) Bft πt sin 0.65 or π t their (b) answer ; 5 5 t awrt. ; A Note B : 8.5 or 6 + their R found in part (a) as long as the answer is greater than 6. πt π t : sin ± theirα or their (b) answer 5 5 A: For sin (0.) This can be implied by awrt. or awrt.0. (d) π t 6 +.5sin ; 5 π t sin ; 5.5 π t 0.65 sin (0.) or awrt 0. A 5 Either t awrt. or awrt 6.7 A π t dd 5 Times : 06, 8: A 6 c So, 0.65 { π or } { }

3 Note πt : 6 + (their R) sin ± theirα 5 πt ± theirα 5 their R 7, : A: For sin (0.). This can be implied by awrt 0. or awrt.7 or other values for different α s. Note this mark can be implied by seeing.055. A: Either t awrt. or t awrt 6.7 dd: either π their PV c. Note that this mark is dependent upon the two M marks. This mark will usually be awarded for seeing either.70 or.7 A: Both t : 06 and t 8 : or both 6 (min) and 0 (min) or both hr 6 min and 6 hr min. [5]. (a) π 5cosx sin x R cos(x + α), R > 0, 0 < x< 5cosx sinx Rcos x cos α R sin x sin α Equate cos x: 5 R cos α Equate sin x: R sin α { } R 5 + ; R 5 + ; or awrt 5.8 A tan α c 5 α tan α ± or tanα ± or 5 5 Hence, 5cos x sin x cos( x ) sin α ± or their R 5 cos α ± or their R α awrt 0.5 or α awrt 0.7π or π or α awrt 5.8 A

4 (b) 5cos x sin x ( x ) cos cos(x ) { } cos(x ± their α) their R (x ) c For applying cos their R x c awrt 0.7 c A (x ) π c { c } π their 0.88 dd x c awrt.9 c A 5 Hence, x {0.7,.9} Note If there are any EXTRA solutions inside the range 0 x < π, then withhold the final accuracy mark if the candidate would otherwise score all 5 marks. Also ignore EXTRA solutions outside the range 0 x < π. [9]. cosec x cot x, (eqn *) 0 x 80 Using cosec x + cot x gives Writing down or using cosec x ± ± cot x + cot x cot x or cosec θ ± ± cot θ cot x cot x 0 or cot x cot x cot x(cot x ) 0 or cot x For either cot x cot x{0} or cot x cot x Attempt to factorise or solve a quadratic (See rules for factorising quadratics) or cancelling out cot x from both sides. d cot x 0 or cot x Both cot x 0 and cot x. A A

5 cot x 0 (tan x ) x 90, 70 Candidate attempts to divide at least one of their principal anglesby. This will be usually implied by seeingx.5 resulting from cot x. dd x 5, 5 cot x tan x x 5, 5 x.5,.5 Overall, x {.5, 5,.5, 5} Both x.5 and x.5 A Both x 5 and x 5 B If there are any EXTRA solutions inside the range 0 x 80 and the candidate would otherwise score FULL MARKS then withhold the final accuracy mark (the sixth mark in this question). Also ignore EXTRA solutions outside the range 0 x 80. [7] 5. (a) cos θ +sin θ ( cos θ) cos cos θ sin θ + θ cos θ cos θ Dividing cos θ +sin θ by cos θ to give underlined equation. + tan θ sec θ tan θ sec θ (as required) AG Complete proof. A cso No errors seen.

6 (b) tan θ + secθ + sec θ, (eqn *) 0 θ< 60 Substituting tan θ sec θ (sec θ ) + secθ + sec θ into eqn * to get a quadratic in secθ only sec θ + secθ + sec θ sec θ + secθ 0 Forming a three term one sided quadratic expression in secθ. (secθ + )(secθ ) 0 Attempt to factorise or solve a quadratic. secθ or secθ cosθ or cosθ cosθ ; or cos θ cosθ ; A; α 0 or α no solutions θ 0 0 A θ 0 0 or θ 60 θ when Bft 6 solving using cosθ... θ {0, 0 } Note the final A mark has been changed to a B mark. [8] 6. (a) A B cos (A + A) cosa cosacosa sinasina Applies A B to cos(a + B) to give the underlined equation or cos A cos A sin A cos A cos A sin A and cos A + sin A gives cosa sin A sin A sin A (as required) Complete proof, with a link between LHS and RHS. A AG No errors seen.

7 (b) C C sin x sin x cos x Eliminating y correctly. Using result in part (a) to substitute for sin x as cos x ± ± cos x sin x cosx or ksin x as ± ± cos x k to produce an equation in only double angles. sin x ( cosx) cosx sin x cosx cosx sin x + cosx Rearranges to give correct result A AG (c) sin x + cos x R cos(x α) sinx + cosx Rcosxcosα + Rsinxsinα Equate sin x: R sin α Equate cos x: R cos α R + ; 5 5 R 5 B tan α º sin α ± or tan α ± or sin α ± or cosα ± their R their R awrt 6.87 A Hence, sin x + cosx 5cos(x 6.87)

8 (d) sin x + cosx 5cos(x 6.87) cos(x 6.87) 5 cos(x ± their α) their R (x 6.87) awrt 66 A (x 6.87) Hence, x o, One of either awrt 5.6 or awrt 5.7 or awrt A 65. or awrt 65. Both awrt 5.6 AND awrt 65. A If there are any EXTRA solutions inside the range 0 x < 80 then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 x <80. [] 7. (a) sin x sin x cos x B aef

9 (b) cosec x 8 cos x 0, 0 < x < π 8cos x 0 sin x Using cosec x sin x 8cos x sin x 8 sin x cos x ( sin x cos x) sin x sin x sin x k, where < k < and k 0 sin x A Radians x { , } Degrees x { , } Either arwt 7. or 8.76 or 0. Radians x {0.6...,.5...} or. or.5 or awrt 0.0π or A Degrees x { , } awrt 0.6π. Both 0. and. A cao 5 Solutions for the final two A marks must be given in x only. If there are any EXTRA solutions inside the range 0 < x < π then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < x < π. [6] 8. (a) (i) sin θ sin(θ + θ) sin θ cosθ + cos θ sin θ sinθ cosθ.cosθ + ( sin θ)sin θ A sinθ ( sinθ) + sinθ sin θ sinθ sin θ cso A

10 (ii) 8sin θ 6sin θ + 0 sinθ + 0 sin θ A π 5π θ, 6 6 π 5π θ, 8 6 A A 5 (b) sin5 sin(60 5 ) sin 60 cos 5 cos 60 sin 5 A 6 ( 6 ) cso A Alternatives sin5 sin (5 0 ) sin 5 cos0 cos 5 sin 0 A 6 ( 6 ) cso A Using cos θ sin θ, cos0 sin 5 sin 5 cos0 sin 5 A ( 6 ) 6 (6 + ) Hence sin5 ( 6 ) cso A []

11 9. (a) R + R 5 A tan α α 5... awrt 5 A (b) Maximum value is 5 ft their R B ft At the maximum, cos(θ α) or θ α 0 θ α 5... ft their α A ft (c) f(t) 0 + 5cos(5t α) Minimum occurs when cos (5t α) The minimum temperature is (0 5) 5 A ft (d) 5t α 80 t 5.5 awrt 5.5 A [] 0. (a) R 5 + R A tan α 5 α.76 cao A (b) cos(x α) 6 x α arccos 6.09 A x awrt. A x α.09 accept 5.9 for M x awrt 0.08 or A 5 (c) (i) R max ft their R Bft (ii) At the maximum, cos (x α) or x α 0 x α.76 awrt., ft their α Aft []

12 . (a) sin θ + cos θ sin sin θ cos θ θ + sin θ sin θ sin θ + cot θ cosec θ * cso A Alternative + cot cos θ sin θ + cos θ θ + sin θ sin θ sin θ cosec θ * cso A (b) (cosec θ ) 9 cosec θ cosec θ 9 cosec θ 5 0 or 5 sin θ + 9 sin θ 0 ( cosec θ + ) (cosec θ 5) 0 or (5 sin θ ) (sin θ + ) 0 cosec θ 5 or sin θ 5 A θ.5, 68.5 AA 6 [8]. (a) cos (x + x) cos x cos x sin x sin x ( cos x ) cos x ( sin x cos x) sin x ( cos x ) cos x ( cos x) cos x any correct expression A cos x cosx A (b) (i) cos x + sin x cos x + ( + sin x) + + sin x cos x ( + sin x) cos x cos x + + sin x + sin x A ( + sin x) cos x ( + sin x) ( + sin x) cos x sec x (*) cos x cso A (c) sec x or cos x x π 5π, accept awrt.05, 5. A, A []

13 . (a) Complete method for R: e.g. R cos α, R sin α, R ( + ) R or.6 (or more accurate) A Complete method for tan α [Allow tan α ] α (Allow.7 ) A st for correct method for R nd for correct method for tanα No working at all: A for, A for or.7. N.B. R cos α, R sin α used, can still score A for R, but loses the A mark for α. cosα, sin α : apply the same marking. (b) Greatest value ( ) 69, A for realising sin(x + α) ±, so finding R. (c) sin(x ) ( ) sin(x + their α) their R (x ) 0.8 (0...) or 6. ) A (x ) π Must be π their 0.8 or 80 their 6. or (x ) π Must be π + their 0.8 or 60 + their 6. x.7 or x (awrt) Both (radian only) A 5 If 0.8 or 6. not seen, correct answers imply this A mark Alt (i) Squaring to form quadratic in sin x or cos x [ cos x cosx 8 0, sin x 6sinx 0] Correct values for cos x ; or sinx 0.767,.7 awrt A For any one value of cos x or sin x, correct method for two values of x x.7 or x (awrt) Both seen anywhere A Checking other values (0.07,.0 or 0.869,.9) and discarding

14 Alt (ii) Squaring and forming equation of form a cosx + bsinx c 9sin x + cos x + sinx sinx + 5cosx Setting up to solve using R formula e.g. cos(x.76) (x.76) cos 0.56(0... (α) A (x.76) π α, π + α,... x.7 or x (awrt) Both seen anywhere A Checking other values and discarding []. (a) sinθ cosθ sin θ + cos θ + cosθ sinθ cosθ sinθ Use of common denominator to obtain single fraction cosθ sinθ Use of appropriate trig identity (in this case sin θ + cos θ ) Use of sin θ sinθ cosθ sin θ cosecθ (*) Acso Alt. (a) sinθ cosθ tan θ + + tanθ + cosθ sin θ tanθ tanθ sec tanθ cosθ sinθ sin θ cosecθ (*) (cso) A If show two expressions are equal, need conclusion such as QED, tick true.

15 (b) y O 90º 80º 70º 60º Shape (May be translated but need to see sections ) T.P.s at y ±, asymptotic at correct x-values (dotted lines not required) Bdep B (c) cosecθ sinθ Allow [ for equation in sinθ], A sin θ (θ) [.80..., ; , ] ; st for α, 80 α; nd adding 60 to at least one of values ; θ 0.9, 69., 00.9, 9. ( d.p.) awrt st Afor any two correct, nd A for other two A, A 6 Extra solutions in range lose final A only SC: Final marks: θ 0.9, after M0M0 is B; record as M0M0AA0 Alt. (c) tan θ + and form quadratic, tan θ tanθ + 0 tanθ, A ( for attempt to multiply through by tanθ, A for correct equation above). ± 5 Solving quadratic [tanθ or ] θ 69., 9. θ 0.9, 00.9 ( d.p.), A, A ( is for one use of 80 + α, AA as for main scheme) [] 5. (a) sin θ sin (θ + θ) sin θ cos θ + cos θ sin θ B sin θ cos θ + ( sin θ) sin θ BB sin θ sin θ + sin θ sin θ sin θ sin θ * cso A 5

16 (b) sin θ or exact A equivalent [7] 6. (a) R ( ) + R A tan α α π accept awrt.05 A (b) sin(x + their α) π 5π π x + their α, A π π xi, 6 accept awrt.57, 5.76 A The use of degrees loses only one mark in this question. Penalise the first time it occurs in an answer and then ignore. [8] 7. (i) sec x cosec x ( + tan x) ( + cot x) A tan x cot x * cso A Alternatives sec x tan x cosec x cot x A Rearranging sec x cosec x tan x cot x * cso A sin x cos x LHS cos x sin x cos x sin x sin x cos x sin x cos x (sin x cos x)(sin x + cos x) RHS cos x sin x cos x sin x cos x sin x sin x cos x cos x sin x A LHS * or equivalent A (ii) (a) y arccos x x cos y B π π x sin y arcsin x y B Accept arcsin x arcsin cos y

17 (b) arccos x + arcsin x π π y + y B [6] 8. (a) Dividing sin θ + cos θ by sin θ to give sin θ cos θ + sin θ sin θ sin θ Completion: + cot θ cosec cosec θ cot θ AG A* (b) cosec θ cot θ (cosec θ cot θ)(cosec θ + cot θ) (cosec θ + cot θ) using (a) AG A* Using LHS ( + cot θ) cot θ, using (a) & elim. cot θ, conclusion {using (a) again} A* ( cos θ )( + cos θ ) Conversion to sines and cosines: needs sin θ for (c) Using (b) to form cosec θ + cot θ cotθ Forming quadratic in cotθ + cot θ + cot θ cotθ {using (a)} cot θ + cotθ 0 Solving: ( cot θ )(cotθ + 0) to cot θ cotθ or cot θ A θ 5 (or correct value(s) for candidate dep. on Ms) Aft 6 Ignore solutions outside range Extra solutions in range loses Aft, but candidate may possibly have more than one correct solution. A [0]

18 9. (a) Method for finding sin A 7 sin A A A 7 First A for, exact Second A for sign (even if dec. answer given) Use of sin A sin A cos A sin A 7 or equivalent exact Aft 5 8 ± f.t. Requires exact value, dependent on nd M π π (b) (i) cos x + + cos x π π π π cos x cos sin x sin + cos x cos + sin x sin π cos x cos A [This can be just written down (using factor formulae) for A] cosx AG A* π A earned, if cos x cos just written down, using factor theorem Final A* requires some working after first result. dy (b) (ii) 6sin x cos x sin x B B dx or π π 6sin x cos x sin x + sin x sinx sinx sin x AG A First B for 6 sin x cos x ; second B for remaining term(s) []

19 0. (a) (i) cos x cos x sin x cos x + sin x cos x + sin x (cos x + sin x)(cos x sin x) cos x + sin x cos x sin x AG A (ii) (cos x sin x) ( cos x sin x cos x), cos x sin x cos x AG A (b) cos θ cos θ cosθ + sinθ cosθ (cosθ sinθ) cos θ cosθ sinθ (cos θ + ) sin θ (cos θ sin θ) 0 sin θ cos θ AG A (c) sin θ cos θ tan θ θ π, 5π, 9π, π A for θ π, 5π, 9π, π ( solns) A []. (a) (i) d (e x+ ) e x+ (or e e x ) At any stage B dx dy x e x+ + xe x+ Or equivalent A+A dx

20 (ii) d (cos(x )) 6x sin(x ) At any stage A dx dy dx 8x sin(x 9x ) cos(x Alternatively using the product rule for second A y (x) cos(x ) dy (x) cos(x ) 6x (x) sin(x ) dx Accept equivalent unsimplified forms ) A dy (b) 8 cos(y + 6) dx or dx dy 8cos(y + 6) dy dx 8cos(y + 6) A dy dx ( ± ) 8cos arcsin x (6 x ) A 5 []. (a) R cos α, R sin α R ( + ) 60 Accept if just written down, awrt.6 A tan α, α 8. awrt 8., A (b) cos (x + their α) 7 ( 0.55) their R x + their α 56. awrt 56 A..., their principal value x 8.0, 85. Ignore solutions out of range A, A 5 If answers given to more than dp, penalise first time then accept awrt above. (c) (i) minimum value is 60 ft their R Bft

21 (ii) cos (x + their α) x 6.57 cao A []. (a) sin(θ + 0) cos(θ + 60) sinθ cos0 + cosθ sin0 cosθ cos60 sinθ sin60 BB sinθ + cosθ cosθ sinθ Finding tanθ, tanθ or equiva. Exact,A 5 (b) (i) Setting A B to give cos A cos A sin A Correct completion: ( sin A) sin A sin A A Need to see intermediate step above for A (ii) Forming quadratic in sinx [ sin x + sin x 0] Solving [( sin x )(sinx + ) 0 or formula] [sin θ ½ or sin θ ] π 5π θ, ; 6 6 A,Aft θ π Aft for π α A 5 (iii) LHS siny cosy sin y cos y B use of tany cosy only + ( sin y) B sin y, forming expression in siny, cos y Completion: sin y + ( sin y) AG A sin y sin y + cos y cos y [Alternative: LHS B cos y cos( y y) A] cos y [5]

22 . (a) Dividing by cos sin θ cos θ θ: + cos θ cos θ cos θ Completion: + tan θ sec θ A (no errors seen) (b) use of + tan θ sec θ: (sec θ ) + secθ [sec θ + secθ 0] Factorising or solving: ( secθ + )(secθ ) 0 [secθ or secθ ] θ 0 B cos θ ; θ.8 A θ 8. A ft 6 [Aft for θ 60 θ ] [8] 5. (a) cos A cos A sin A (+ use of cos A + sin A ) ( sin A); sin A sin A (*) A (b) sin θ cos θ sinθ + sinθ; ( sin θ) sinθ + B; sinθ cosθ + 6sin θ sinθ sinθ(cosθ + 6sinθ ) (*) A (c) cosθ + 6sinθ Rsinθcosα + Rcosθ sinα Complete method for R (may be implied by correct answer) [R + 6, Rsinα, Rcosα 6] R 5 or 7. A Complete method for α; α A (allow.7 ) (d) sinθ(cosθ + 6sinθ ) 0 θ 0 B sin(θ ) (.6 ) 5 θ (0.9),.75 [or θ +.7 (.6 ), 55. ] d θ. cao A 5 [5]

23 6. (a) Complete method for DE [e.g. split triangle ADE and sin, or sine or cos rule] DE sin θ (*) ( c.s.o.) A(*) (b) P DE + EF or equivalent. With attempt at EF 8sin θ + cos θ (*) ( c.s.o.) A(*) (c) 8sin θ + cos θ R sin (θ + α) R sin θ cos α + R cos θ sin α Method for R, method for α need to use tan for nd M [R cos α 8, R sin α tan α 0.5, R ( 8 + ] R 5 or 8.9, α 0.6 (allow 6.6), A A awrt 0.6 (d) Using candidate s R sin (θ + α) 8.5 to give (θ + α) sin 8.5 R Solving to give θ sin 8.5 R α, θ 0.79 (allow 5.) A Considering second angle: θ + α π (or 80) sin 8.5 ; R θ. (allow 8.6) A 5 [] 7. (i) A correct form of cos x used 7 or or A 5 sec x ; or cos x 7 7 A

24 (ii) cos x (a) + or (b) + sin x sin x tan x sin x Forming single fraction (or ** multiplying both sides by sinx) Use of correct trig. formulae throughout and producing expression in terms of sinx and cosx Completion (cso) e.g. cos x sin x cos x cos x cot x (*) A sin x [8] 8. (a) (i) cosθ 5sinθ Rcosθcosα Rsinθsinα. R 5 α R 5 +, R 5 tan α, α.6 (AWRT.6), A or 0.9 C (AWRT 0.9 C ), A for correct expression for R or R for correct trig expression for α (b) (i) cos(θ +.6) θ , θ 9.5 (only) A cos(θ + α) R θ + α ft their R

25 (ii) 8 tanθ tanθ i.e. 0 tan θ + tanθ 8 0 (tanθ )(tanθ + ) tanθ or tanθ θ 5. A [ignore θ not in range e.g. θ 6.6] A 5 Use of cot θ tanθ TQ in tan θ 0 Attempt to solve TQ 0 A For Final A mark must deal with tan θ [] 9. (a) sin θ sin θ tan θ cos θ or cos θ or equivalent + tan sin θ θ sec θ + cos θ cos θ sin θ cos θ cosθ (*) cos θ + sin θ A cso

26 π (b) θ, cosθ 8 t + t t +. + (*) A 5 cso Alternative to (b) t t tanθ t + t 0 t t ( ) (*) A 5 cso [9] 0. (i) cos x cos 0 sin x sin 0 (cos x cos 0 + sin x sin 0) Correct use of cos(x ± 0) cos x sin x cos x + sin x Sub. for sin 0 etc decimals, surds A, A i.e. sin x cos x tan x (*), Acso 5 sin x Collect terms and use tan x cos x (ii) (a) ( sin θ ) LHS sinθ cosθ Use of cos A or sin A; both correct ; A sinθ tan θ (*) A cso cosθ (b) Verifying: 0 (since sin 60 0, cos 60 ) B cso

27 ( cos θ ) (c) Equation sin θ Rearrange to form tan θ or cot θ cos θ sin θ A i.e. θ (6.6 or 06.6 ) AWRT 7, 07 st solution must be tanθ ± or (both) A [] Alt (c) sin θ cos θ ( sin θ) Use of cos A and sin A 0 sin θ ( sin θ cos θ) (sin θ 0) tan θ etc, as in scheme A Alt (c) cos θ + sin θ cos (θ α) α.6 (or 7) A 5 θ α, 60, 60 + α θ α, 80 + α i.e. θ 7 or 07 (or dp) θ α or 80 + α A both. (a) sin x + cos x R sin (x + α) R (sin x cos α + cos x sin α) R cos α, R sin α Method for R or α, e.g. R ( + ) or tan α A Both R and α 60 A

28 (b) sec x + cosec x + cos x sin x + cos x sin x cos x sin x B sin x (*) (c) Clearly producing sin x sin (x + 60) A (d) sin x sin (x + 60) 0 cos cos sin x + 60 sin x 60 0 x x 0, 60 A A ft x 60 0 x 60 B 5 []. (a) y (, ) O x (a) (, ) y arcsin x Shape correct passing through O: G; end-points: G

29 (, ) y (, ) O x (b) y sec x Shape correct, symmetry in Oy: G end-points: G (c) π π π π x sec x Area estimate π π π + sec x dx A A.78 ( d.p.) A [8]. (a) LHS sin x + sin 60 cos 60 sin x sin x + cos x sin x A RHS sin 60 cos x + cos 60 sin x cos x + sin x LHS A (b) From (a), sin (60 + x) sin (60 x) sin x x sin 8 sin 6 sin α A

30 (c) sin x + sin x + sin (60 x) sin (60 + x) Using (a), sin x A x 99.7 or 0.5 x 99.7, 70. A or 79.7, 50. A ft 5 []. (a) e w 0.5 w.9 A (b) arctan x x 0.79 A (c) ln y + y 0.85 y + y e 0.85 A.0 y 0.75 A y (d) π Putting cos z + sin z in the form cos z or equivalent A π cos z attempt for z z., 0.55 A, A ft 5 [] 5. (a) (i) sin (A + B) sin (A B) sin A cos B + sin B cos A sin A cos B + sin B cos A sin B cos A (*) A cso (ii) cos (A B) cos (A + B) cos A cos B + sin A sin B cos A cos B + sin A sin B sin A sin B (*) A cso (b) sin cos ( A + B) sin( A B) ( A B) sin( A + B) sin B cos A sin A sin B cos A sin A A cot A (*) A cso

31 (c) Let A 75 and B 5 B sin 90 cos 60 o o sin 60 o cos 90 o cot 75 o cot 75 0 A [] 6. (a) sin θ cos θ R sin θ cos α R cos θ sin α sin θ terms give R cos α cos θ terms give R sin α tan α 0.75 α 6.9 o A R + 5 R 5 A (b) 5 sin (θ 6.9 o ) sin (θ 6.9 o ) 0.6 θ 6.9 o 6.9 o,. θ 7.7 o, 80 awrt 7 A A A5 (c) Max value 5 B (d) sin (θ 6.9 o ) θ θ A [] (c) Shape correct, symmetry in Oy: G end-points: G π π π π π π + x 6 6 sec x dx π

Cambridge International Examinations Cambridge International General Certificate of Secondary Education. Published

Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/ Paper May/June 07 MARK SCHEME Maximum Mark: 80 Published This mark scheme