OCR C3 2011 June — Question 9 12 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeProve identity then solve equation only (no integral)
DifficultyStandard +0.3 Part (i) is a standard compound angle formula proof requiring systematic expansion and simplification. Part (ii) directly applies the proven identity with pattern recognition. Part (iii) requires recognizing the identity again and solving a simple linear equation in 6θ. This is a well-structured multi-part question testing routine application of addition formulae with no novel insights required, making it slightly easier than the average A-level question.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities

9
  1. Prove that \(\frac { \sin ( \theta - \alpha ) + 3 \sin \theta + \sin ( \theta + \alpha ) } { \cos ( \theta - \alpha ) + 3 \cos \theta + \cos ( \theta + \alpha ) } \equiv \tan \theta\) for all values of \(\alpha\).
  2. Find the exact value of \(\frac { 4 \sin 149 ^ { \circ } + 12 \sin 150 ^ { \circ } + 4 \sin 151 ^ { \circ } } { 3 \cos 149 ^ { \circ } + 9 \cos 150 ^ { \circ } + 3 \cos 151 ^ { \circ } }\).
  3. It is given that \(k\) is a positive constant. Solve, for \(0 ^ { \circ } < \theta < 60 ^ { \circ }\) and in terms of \(k\), the equation $$\frac { \sin \left( 6 \theta - 15 ^ { \circ } \right) + 3 \sin 6 \theta + \sin \left( 6 \theta + 15 ^ { \circ } \right) } { \cos \left( 6 \theta - 15 ^ { \circ } \right) + 3 \cos 6 \theta + \cos \left( 6 \theta + 15 ^ { \circ } \right) } = k .$$

AnswerMarks Guidance
(i) Use at least one identity correctlyB1 angle-sum or angle-difference identity
Attempt use of relevant identities in single rational expressionM1 not earned if identities used in expression where step equiv to \(\frac{A + B + C}{D + E + F} = \frac{A}{D} + \frac{B}{E} + \frac{C}{F}\) or similar has been carried out; condone (for M1A1) if signs of identities apparently switched (so that, for example, denominator appears as \(\cos \theta \cos \alpha - \sin \theta \sin \alpha +\) \(3\cos \theta + \cos \theta \cos \alpha + \sin \theta \sin \alpha\))
Obtain \(\frac{2\sin \theta \cos \alpha + 3\sin \theta}{2\cos \theta \cos \alpha + 3\cos \theta}\)A1 or equiv but with the other two terms from each of num'r and den'r absent
Attempt factorisation of num'r and den'rM1
Obtain \(\frac{\sin \theta}{\cos \theta}\) and hence \(\tan \theta\)A1 5 marks: AG; necessary detail needed
(ii) State or imply form \(k \tan 150°\)M1 obtained without any wrong method seen
State \(\frac{1}{3}\tan 150°\)A1 or equiv such as \(\frac{12\sin 150°}{9\cos 150°}\)
Obtain \(-\frac{4}{3}\sqrt{3}\)A1 3 marks: or exact equiv (such as \(-\frac{-4\sqrt{3}}{3}\)); correct answer only earns 3/3
(iii) State or imply \(\tan 6\theta = k\)B1
State \(\frac{1}{6}\tan^{-1}k\)B1
Attempt second value of \(\theta\)M1 using \(6\theta = \tan^{-1}k + (\)multiple of 180)
Obtain \(\frac{1}{6}\tan^{-1}k + 30°\)A1 4 marks: and no other value
**(i)** Use at least one identity correctly | B1 | angle-sum or angle-difference identity

Attempt use of relevant identities in single rational expression | M1 | not earned if identities used in expression where step equiv to $\frac{A + B + C}{D + E + F} = \frac{A}{D} + \frac{B}{E} + \frac{C}{F}$ or similar has been carried out; condone (for M1A1) if signs of identities apparently switched (so that, for example, denominator appears as $\cos \theta \cos \alpha - \sin \theta \sin \alpha +$ $3\cos \theta + \cos \theta \cos \alpha + \sin \theta \sin \alpha$)

Obtain $\frac{2\sin \theta \cos \alpha + 3\sin \theta}{2\cos \theta \cos \alpha + 3\cos \theta}$ | A1 | or equiv but with the other two terms from each of num'r and den'r absent

Attempt factorisation of num'r and den'r | M1 |

Obtain $\frac{\sin \theta}{\cos \theta}$ and hence $\tan \theta$ | A1 | 5 marks: AG; necessary detail needed

**(ii)** State or imply form $k \tan 150°$ | M1 | obtained without any wrong method seen

State $\frac{1}{3}\tan 150°$ | A1 | or equiv such as $\frac{12\sin 150°}{9\cos 150°}$

Obtain $-\frac{4}{3}\sqrt{3}$ | A1 | 3 marks: or exact equiv (such as $-\frac{-4\sqrt{3}}{3}$); correct answer only earns 3/3

**(iii)** State or imply $\tan 6\theta = k$ | B1 |

State $\frac{1}{6}\tan^{-1}k$ | B1 |

Attempt second value of $\theta$ | M1 | using $6\theta = \tan^{-1}k + ($multiple of 180)

Obtain $\frac{1}{6}\tan^{-1}k + 30°$ | A1 | 4 marks: and no other value

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9 (i) Prove that $\frac { \sin ( \theta - \alpha ) + 3 \sin \theta + \sin ( \theta + \alpha ) } { \cos ( \theta - \alpha ) + 3 \cos \theta + \cos ( \theta + \alpha ) } \equiv \tan \theta$ for all values of $\alpha$.\\
(ii) Find the exact value of $\frac { 4 \sin 149 ^ { \circ } + 12 \sin 150 ^ { \circ } + 4 \sin 151 ^ { \circ } } { 3 \cos 149 ^ { \circ } + 9 \cos 150 ^ { \circ } + 3 \cos 151 ^ { \circ } }$.\\
(iii) It is given that $k$ is a positive constant. Solve, for $0 ^ { \circ } < \theta < 60 ^ { \circ }$ and in terms of $k$, the equation

$$\frac { \sin \left( 6 \theta - 15 ^ { \circ } \right) + 3 \sin 6 \theta + \sin \left( 6 \theta + 15 ^ { \circ } \right) } { \cos \left( 6 \theta - 15 ^ { \circ } \right) + 3 \cos 6 \theta + \cos \left( 6 \theta + 15 ^ { \circ } \right) } = k .$$

\hfill \mbox{\textit{OCR C3 2011 Q9 [12]}}