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1 cos θ + 4sin θ = R cos(θ α) = R (c os θ cos α + sin θ sin α) R cos α =, R sin α = 4 R = + 4 = 5, R = 5 tan α = 4/ α = 0.97 f(θ) = 7 + 5cos(θ 0.97) R=5 tan α=4/ oe ft their R 0.9 or 5. or better their cos (θ ) = or - used (condone use of graphical calculator) and seen cao Range is to simplified Greatest value of 7+ cosθ+4sinθ is ½. ft [6] sin x +cos x = R sin(x + α) = R sin x cos α + R cos x sin α R cos α =, R sin α = R = + =, R = tan α = /, α = sin x cos x sin( x ) maximum when x = / x = / = 0.98 rads y = =.6 So coords of max point are (0.98,.6) [7] Correct pairs. Condone omission of R if used correctly. Condone sign error. or.6 or better, not ± unless+ chosen ft from first or better (accept 0.59), with no errors seen in method for angle (allow.7º or better) any valid method eg differentiating 0.98 only. Do not accept degrees or multiples of π. condone, ft their R if,say = 4
2 LHS = sin cos cos sin cos cos sin tan RHS cos one correct double angle formula used cancelling cos s 4(i) sin(θ + 45 ) = cos θ sin θ cos 45 + cos θ sin 45 = cos θ (/ ) sin θ + (/ ) cos θ = cos θ sin θ + cos θ = cos θ sin θ = ( ) cos θ sin θ = tan θ = * cosθ [5] compound angle formula sin 45 = /, cos 45 = / collecting terms (ii) tan θ = θ =.5, 0.5 [] and no others in the range 5(i) sin θ + cos θ = 4sin θ cos θ + sin θ = sin θ (cos θ sin θ) = 0 or 4 tanθ-tan²θ=0 sin θ = 0 or tan θ =0, θ = 0, 80 or c θ sin θ = 0 tan θ = θ = 6.4, 4.4 OR Using Rsin(θ+α) R= 5 and α=6.57 θ +6.57=arcsin /R θ=0,80 θ=6.4, 4.4, [6], [6] Using double angle formulae Correct simplification to factorisable or other form that leads to solutions 0 and 80 tan θ = (- for extra solutions in range) (- for extra solutions in range)
3 Question Answer Marks Guidance 6 cos sin * cos sin (maybe implied in substitution) (6cos sin ) 6 sin sin 6cos + sin = 0 sin sin 6 = 0 (4sin )(sin + ) = 0 dep* Use of correct quadratic equation formula or factorising or comp. the square on their three term quadratic in sin (see guidance in question for awarding this method mark) provided b 4ac 0 sin = /4 or / www First correct solution to dp or better (eg etc) Three correct solutions sin = /4, = 48.6,.4 sin = /, =.8, 8. All four correct solutions and no others in the range Ignore solutions outside the range SC Award max B0 for answers in radians (0.85,.9,.87, 5.55 or better so one correct, three correct ). Award max if there are extra solutions in the range with radians [7] SC If awarded and both values of sin but B0B0B0 then award only for evidence of using sin sin 80
4 Question Answer Marks Guidance 7 (i) EITHER Use of cos = /sec (or sin= /cosec) From RHS tan tan sec sec sin / cos.sin / cos / cos./ cos cos cos ( sin sin ) cos cos cos cos sin sin Must be used Substituting and simplifying as far as having no fractions within a fraction tt [need more than secsec cc ss ie an intermediate step that can lead to cc ss] cos( ) Convincing simplification and correct use of cos(α + β) Answer given OR From LHS, cos = /sec or sin = /cosec used cos( ) cos cos sin sin sec sec sin sin sec sin sec sin sec sec Correct angle formula and substitution and simplification to one term OR eg cosαcosβ sinαsinβ cos cos ( tan tan ) tan tan sec sec Simplifying to final answer www Answer given Or any equivalent work but must have more than cc ss = answer.
5 Question Answer Marks Guidance 7 (ii) cos tan sec tan tan. β = α used, Need to see sec²α Use of sec²α = + tan²α to give required result Answer Given OR, without Hence, cos cos ( sin ) cos ( tan ) sec tan tan Use of cosα = cos²α sin²α soi Simplifying and using sec²α = + tan²α to final answer Answer Given Accept working in reverse to show RHS=LHS, or showing equivalent 7 (iii) cos = ½ = 60, 00 = 0, 50 [] Soi or from tan²θ = / oe from sin²θ or cos²θ First correct solution Second correct solution and no others in the range SC for π/6and 5π/6 and no others in the range
6 8 (i) BAC = 0 90 (90 θ) = θ 60 BC = b sin(θ 60) CD = AE = a sin θ h = BC + CD = a sin θ + b sin (θ 60 ) * (ii) h = a sin θ + b sin (θ 60 ) = a sin θ + b (sinθ cos 60 cosθ sin 60) = a sin θ + ½ b sin θ / b cos θ = ( a+ b)sin θ bcos θ * corr compound angle formula sin 60 = /, cos 60 = ½ used (iii) OB horizontal when h = 0 ( a+ b) sinθ bcosθ = 0 (a+ b)sin θ = bco s θ b sinθ cosθ = a+ b tan θ = b a+ b * sinθ cosθ = tanθ (iv) sinθ cosθ = R sin(θ α) = R(sinθ cosα cosθ sin α ) R cos α =, R sin α = R = + ( ) = 7, R = 7 =.646 m tan α = /, α = 40.9 So h = 7 sin(θ 40.9 ) h max = 7 =.646 m when θ 40.9 = 90 θ = 0.9 ft [7]
7 9(i) cosθ + sinθ =r cos( θ α) = Rcosθ cosα + Rsinθsin α Rcosα =, Rsinα = R = + ( ) = 4, R = tan α = α = π/ [4] R = equating correct pairs tan α = o.e. (ii) derivative of tan θ is sec θ π sec ( ) dθ (cosθ + sinθ) 4 π π dθ = θ 0 0 π = tan( θ ) 4 = ¼ (0 ( )) = /4 * π 0 [4] ft their α R www [tan (θ π/] ft their R,α(in radians)
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