OCR C3 2011 June — Question 3 8 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2011
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeGiven sin/cos/tan, find other expressions
DifficultyStandard +0.3 This is a straightforward application of double angle formulae with algebraic manipulation. Part (a) requires expanding sin 2α, factoring, and solving; part (b) uses cos 2β = 2cos²β - 1 to form a quadratic. Both are standard C3 exercises requiring routine technique rather than insight, making them slightly easier than average.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae

3
  1. Given that \(7 \sin 2 \alpha = 3 \sin \alpha\), where \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), find the exact value of \(\cos \alpha\).
  2. Given that \(3 \cos 2 \beta + 19 \cos \beta + 13 = 0\), where \(90 ^ { \circ } < \beta < 180 ^ { \circ }\), find the exact value of \(\sec \beta\).

AnswerMarks Guidance
(a) State \(14\sin \alpha \cos \alpha = 3\sin \alpha\)B1 or unsimplified equiv such as \(7(2\sin \alpha \cos \alpha) = 3\sin \alpha\)
Attempt to find value of \(\cos \alpha\)M1 by valid process; may be implied
Obtain \(\frac{3}{14}\)A1 3 marks: exact answer required; ignore subsequent work to find angle
(b) Attempt use of identity for \(\cos 2\beta\)M1 of form \(\pm 2\cos^2 \beta \pm 1\); initial use of \(\cos^2 \beta - \sin^2 \beta\) needs attempt to express \(\sin^2 \beta\) in terms of \(\cos^2 \beta\) to earn M1
Obtain \(6\cos^2 \beta + 19\cos \beta + 10\)A1 or unsimplified equiv or equiv involving \(\sec \beta\)
Attempt solution of 3-term quadratic eqnM1 for \(\cos \beta\) (or after adjustment) for \(\sec \beta\)
Use \(\sec \beta = \frac{1}{\cos \beta}\) at some stageM1 or equiv
Obtain \(-\frac{3}{2}\)A1 5 marks: or equiv; and (finally) no other answer
**(a)** State $14\sin \alpha \cos \alpha = 3\sin \alpha$ | B1 | or unsimplified equiv such as $7(2\sin \alpha \cos \alpha) = 3\sin \alpha$

Attempt to find value of $\cos \alpha$ | M1 | by valid process; may be implied

Obtain $\frac{3}{14}$ | A1 | 3 marks: exact answer required; ignore subsequent work to find angle

**(b)** Attempt use of identity for $\cos 2\beta$ | M1 | of form $\pm 2\cos^2 \beta \pm 1$; initial use of $\cos^2 \beta - \sin^2 \beta$ needs attempt to express $\sin^2 \beta$ in terms of $\cos^2 \beta$ to earn M1

Obtain $6\cos^2 \beta + 19\cos \beta + 10$ | A1 | or unsimplified equiv or equiv involving $\sec \beta$

Attempt solution of 3-term quadratic eqn | M1 | for $\cos \beta$ (or after adjustment) for $\sec \beta$

Use $\sec \beta = \frac{1}{\cos \beta}$ at some stage | M1 | or equiv

Obtain $-\frac{3}{2}$ | A1 | 5 marks: or equiv; and (finally) no other answer

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3
\begin{enumerate}[label=(\alph*)]
\item Given that $7 \sin 2 \alpha = 3 \sin \alpha$, where $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, find the exact value of $\cos \alpha$.
\item Given that $3 \cos 2 \beta + 19 \cos \beta + 13 = 0$, where $90 ^ { \circ } < \beta < 180 ^ { \circ }$, find the exact value of $\sec \beta$.
\end{enumerate}

\hfill \mbox{\textit{OCR C3 2011 Q3 [8]}}