OCR C3 2015 June — Question 9 13 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeProve identity then solve equation only (no integral)
DifficultyStandard +0.8 This is a substantial multi-part question requiring addition formulae expansion, algebraic manipulation to prove identities involving double angle formulae and quartic expressions, then applying these results to find ranges and solve a complex trigonometric equation. While the techniques are all C3 standard, the length, multiple steps, and need to connect parts (i) and (ii) together make this significantly harder than average, though not requiring truly novel insight.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

9 It is given that \(\mathrm { f } ( \theta ) = \sin \left( \theta + 30 ^ { \circ } \right) + \cos \left( \theta + 60 ^ { \circ } \right)\).
  1. Show that \(\mathrm { f } ( \theta ) = \cos \theta\). Hence show that $$f ( 4 \theta ) + 4 f ( 2 \theta ) \equiv 8 \cos ^ { 4 } \theta - 3 .$$
  2. Hence
    1. determine the greatest and least values of \(\frac { 1 } { \mathrm { f } ( 4 \theta ) + 4 \mathrm { f } ( 2 \theta ) + 7 }\) as \(\theta\) varies,
    2. solve the equation $$\sin \left( 12 \alpha + 30 ^ { \circ } \right) + \cos \left( 12 \alpha + 60 ^ { \circ } \right) + 4 \sin \left( 6 \alpha + 30 ^ { \circ } \right) + 4 \cos \left( 6 \alpha + 60 ^ { \circ } \right) = 1$$ for \(0 ^ { \circ } < \alpha < 60 ^ { \circ }\). \section*{END OF QUESTION PAPER}

Question 9:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
Use at least one addition formula accuratelyM1 Without substituting values for \(\cos 30°\) etc. yet
Obtain \(\cos\theta\)A1 AG; necessary detail needed
State \(\cos 4\theta = 2\cos^2 2\theta - 1\)B1 Or \(\cos 4\theta = \cos^2 2\theta - \sin^2 2\theta\)
Attempt correct use of relevant formulae to express in terms of \(\cos\theta\)M1 Or in terms of \(\cos\theta\) and \(\sin\theta\)
Obtain correct unsimplified expression in terms of \(\cos\theta\) onlyA1 e.g. \(2(2c^2-1)^2 - 1 + 4(2c^2-1)\)
Simplify to confirm \(8\cos^4\theta - 3\)A1 AG; necessary detail needed
[6]
Part (ii)(a):
AnswerMarks Guidance
AnswerMarks Guidance
Obtain \(\frac{1}{12}\)B1
Substitute \(0\) for \(\cos\theta\) in correct expressionM1 No need to specify greatest and least
Obtain \(\frac{1}{4}\)A1
[3]
Part (ii)(b):
AnswerMarks Guidance
AnswerMarks Guidance
State or imply \(8\cos^4(3\alpha) - 3 = 1\)B1 Or \(2\cos^2 6\alpha + 4\cos 6\alpha - 2 = 0\)
Attempt correct method to obtain at least one value of \(\alpha\)M1 Allow for equation of form \(\cos^4(3\alpha) = k\) where \(0 < k < 1\) or for three-term quadratic equation in \(\cos 6\alpha\)
Obtain \(10.9\)A1 Or greater accuracy \(10.921\ldots\) — Answer(s) only: 0/4
Obtain \(49.1\)A1 Or greater accuracy \(49.078\ldots\); and no others between 0 and 60
[4]
# Question 9:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use at least one addition formula accurately | M1 | Without substituting values for $\cos 30°$ etc. yet |
| Obtain $\cos\theta$ | A1 | AG; necessary detail needed |
| State $\cos 4\theta = 2\cos^2 2\theta - 1$ | B1 | Or $\cos 4\theta = \cos^2 2\theta - \sin^2 2\theta$ |
| Attempt correct use of relevant formulae to express in terms of $\cos\theta$ | M1 | Or in terms of $\cos\theta$ and $\sin\theta$ |
| Obtain correct unsimplified expression in terms of $\cos\theta$ only | A1 | e.g. $2(2c^2-1)^2 - 1 + 4(2c^2-1)$ |
| Simplify to confirm $8\cos^4\theta - 3$ | A1 | AG; necessary detail needed |

**[6]**

## Part (ii)(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $\frac{1}{12}$ | B1 | |
| Substitute $0$ for $\cos\theta$ in correct expression | M1 | No need to specify greatest and least |
| Obtain $\frac{1}{4}$ | A1 | |

**[3]**

## Part (ii)(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $8\cos^4(3\alpha) - 3 = 1$ | B1 | Or $2\cos^2 6\alpha + 4\cos 6\alpha - 2 = 0$ |
| Attempt correct method to obtain at least one value of $\alpha$ | M1 | Allow for equation of form $\cos^4(3\alpha) = k$ where $0 < k < 1$ or for three-term quadratic equation in $\cos 6\alpha$ |
| Obtain $10.9$ | A1 | Or greater accuracy $10.921\ldots$ — Answer(s) only: 0/4 |
| Obtain $49.1$ | A1 | Or greater accuracy $49.078\ldots$; and no others between 0 and 60 |

**[4]**
9 It is given that $\mathrm { f } ( \theta ) = \sin \left( \theta + 30 ^ { \circ } \right) + \cos \left( \theta + 60 ^ { \circ } \right)$.\\
(i) Show that $\mathrm { f } ( \theta ) = \cos \theta$. Hence show that

$$f ( 4 \theta ) + 4 f ( 2 \theta ) \equiv 8 \cos ^ { 4 } \theta - 3 .$$

(ii) Hence
\begin{enumerate}[label=(\alph*)]
\item determine the greatest and least values of $\frac { 1 } { \mathrm { f } ( 4 \theta ) + 4 \mathrm { f } ( 2 \theta ) + 7 }$ as $\theta$ varies,
\item solve the equation

$$\sin \left( 12 \alpha + 30 ^ { \circ } \right) + \cos \left( 12 \alpha + 60 ^ { \circ } \right) + 4 \sin \left( 6 \alpha + 30 ^ { \circ } \right) + 4 \cos \left( 6 \alpha + 60 ^ { \circ } \right) = 1$$

for $0 ^ { \circ } < \alpha < 60 ^ { \circ }$.

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR C3 2015 Q9 [13]}}