Question 7 - A-Level Maths

Edexcel C34 2017 June — Question 7 9 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2017
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Convert equation to quadratic form
Difficulty	Standard +0.3 Part (a) is a standard double-angle identity proof requiring recall of cos(2x) = 1 - 2sin²x = 2cos²x - 1. Part (b) requires substituting the proven identity, simplifying to a quadratic in cos(θ), and solving within the given range. This is a routine multi-step question testing standard C3/C4 techniques with no novel insight required, making it slightly easier than average.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

(a) Prove that

$$\frac { 1 - \cos 2 x } { 1 + \cos 2 x } \equiv \tan ^ { 2 } x , \quad x \neq ( 2 n + 1 ) 90 ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$, $$\frac { 2 - 2 \cos 2 \theta } { 1 + \cos 2 \theta } - 2 = 7 \sec \theta$$ Give your answers in degrees to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Working	Marks	Guidance
$\frac{1-\cos 2x}{1+\cos 2x} = \frac{1-(1-2\sin^2 x)}{1+(2\cos^2 x -1)} = \frac{2\sin^2 x}{2\cos^2 x} = \tan^2 x$	M1 dM1 A1	M1: Uses a correct double angle identity on numerator or denominator and applies this to the fraction. dM1: Uses correct double angle identities in numerator and denominator leading to form $\frac{a\sin^2 x}{a\cos^2 x}$. A1*: Completely correct solution; variables must be consistent; do not accept $\frac{\sin^2}{\cos^2} = \tan^2$ within proof.

(3 marks)

Part (b):

Answer	Marks	Guidance
Working	Marks	Guidance
$2\left(\frac{1-\cos 2\theta}{1+\cos 2\theta}\right) - 2 = 7\sec\theta \Rightarrow 2\tan^2\theta - 2 = 7\sec\theta$	M1	M1: Obtains equation of form $A\tan^2\theta - 2 = 7\sec\theta$ or $A\tan^2\theta - 2 = \frac{7}{\cos\theta}$
$\Rightarrow 2(\sec^2\theta - 1) - 2 = 7\sec\theta$	M1	M1: Attempts to use $\tan^2\theta = \pm\sec^2\theta \pm 1$ to produce quadratic in $\sec\theta$
$\Rightarrow 2\sec^2\theta - 7\sec\theta - 4 = 0$	A1	A1: Correct equation. Either $2\sec^2\theta - 7\sec\theta - 4 = 0$ or $4\cos^2\theta + 7\cos\theta - 2 = 0$
$\Rightarrow (2\sec\theta + 1)(\sec\theta - 4) = 0$
$\Rightarrow \sec\theta = -\frac{1}{2}, 4$
$\Rightarrow \cos\theta = -2, \frac{1}{4} \Rightarrow \theta = \ldots$	M1	M1: Correct method of solving 3TQ in $\sec\theta$ or $\cos\theta$ and using arccos to produce at least one answer. If roots are incorrect and no working shown, score M0.
$\Rightarrow \theta = 75.5°, -75.5°$	A1, A1	A1: One of awrt $\theta = 75.5°, -75.5°$. A1: Both. Deduct final mark for extra answers in range. Ignore answers outside range.

(6 marks) — (9 marks total)

## Question 7:

### Part (a):

| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1-\cos 2x}{1+\cos 2x} = \frac{1-(1-2\sin^2 x)}{1+(2\cos^2 x -1)} = \frac{2\sin^2 x}{2\cos^2 x} = \tan^2 x$ | M1 dM1 A1 | M1: Uses a correct double angle identity on numerator **or** denominator **and applies this to the fraction**. dM1: Uses correct double angle identities in numerator and denominator leading to form $\frac{a\sin^2 x}{a\cos^2 x}$. A1*: Completely correct solution; variables must be consistent; do not accept $\frac{\sin^2}{\cos^2} = \tan^2$ within proof. |

**(3 marks)**

### Part (b):

| Working | Marks | Guidance |
|---------|-------|----------|
| $2\left(\frac{1-\cos 2\theta}{1+\cos 2\theta}\right) - 2 = 7\sec\theta \Rightarrow 2\tan^2\theta - 2 = 7\sec\theta$ | M1 | M1: Obtains equation of form $A\tan^2\theta - 2 = 7\sec\theta$ or $A\tan^2\theta - 2 = \frac{7}{\cos\theta}$ |
| $\Rightarrow 2(\sec^2\theta - 1) - 2 = 7\sec\theta$ | M1 | M1: Attempts to use $\tan^2\theta = \pm\sec^2\theta \pm 1$ to produce quadratic in $\sec\theta$ |
| $\Rightarrow 2\sec^2\theta - 7\sec\theta - 4 = 0$ | A1 | A1: Correct equation. Either $2\sec^2\theta - 7\sec\theta - 4 = 0$ or $4\cos^2\theta + 7\cos\theta - 2 = 0$ |
| $\Rightarrow (2\sec\theta + 1)(\sec\theta - 4) = 0$ | | |
| $\Rightarrow \sec\theta = -\frac{1}{2}, 4$ | | |
| $\Rightarrow \cos\theta = -2, \frac{1}{4} \Rightarrow \theta = \ldots$ | M1 | M1: Correct method of solving 3TQ in $\sec\theta$ or $\cos\theta$ **and** using arccos to produce at least one answer. If roots are incorrect and no working shown, score M0. |
| $\Rightarrow \theta = 75.5°, -75.5°$ | A1, A1 | A1: One of awrt $\theta = 75.5°, -75.5°$. A1: Both. Deduct final mark for extra answers in range. Ignore answers outside range. |

**(6 marks) — (9 marks total)**

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Show LaTeX source

\begin{enumerate}
  \item (a) Prove that
\end{enumerate}

$$\frac { 1 - \cos 2 x } { 1 + \cos 2 x } \equiv \tan ^ { 2 } x , \quad x \neq ( 2 n + 1 ) 90 ^ { \circ } , n \in \mathbb { Z }$$

(b) Hence, or otherwise, solve, for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$,

$$\frac { 2 - 2 \cos 2 \theta } { 1 + \cos 2 \theta } - 2 = 7 \sec \theta$$

Give your answers in degrees to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\

\hfill \mbox{\textit{Edexcel C34 2017 Q7 [9]}}

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Q1 6 Q2 6 Q3 8 Q4 8 Q5 8 Q6 6 Q7 9 Q8 7 Q9 8 Q10 9 Q11 11 Q12 14 Q13 9 Q14 16

Answer	Marks	Guidance
Working	Marks	Guidance
\(2\left(\frac{1-\cos 2\theta}{1+\cos 2\theta}\right) - 2 = 7\sec\theta \Rightarrow 2\tan^2\theta - 2 = 7\sec\theta\)	M1	M1: Obtains equation of form \(A\tan^2\theta - 2 = 7\sec\theta\) or \(A\tan^2\theta - 2 = \frac{7}{\cos\theta}\)
\(\Rightarrow 2(\sec^2\theta - 1) - 2 = 7\sec\theta\)	M1	M1: Attempts to use \(\tan^2\theta = \pm\sec^2\theta \pm 1\) to produce quadratic in \(\sec\theta\)
\(\Rightarrow 2\sec^2\theta - 7\sec\theta - 4 = 0\)	A1	A1: Correct equation. Either \(2\sec^2\theta - 7\sec\theta - 4 = 0\) or \(4\cos^2\theta + 7\cos\theta - 2 = 0\)
\(\Rightarrow (2\sec\theta + 1)(\sec\theta - 4) = 0\)
\(\Rightarrow \sec\theta = -\frac{1}{2}, 4\)
\(\Rightarrow \cos\theta = -2, \frac{1}{4} \Rightarrow \theta = \ldots\)	M1	M1: Correct method of solving 3TQ in \(\sec\theta\) or \(\cos\theta\) and using arccos to produce at least one answer. If roots are incorrect and no working shown, score M0.
\(\Rightarrow \theta = 75.5°, -75.5°\)	A1, A1	A1: One of awrt \(\theta = 75.5°, -75.5°\). A1: Both. Deduct final mark for extra answers in range. Ignore answers outside range.