Question 5 - A-Level Maths

Edexcel C3 2012 June — Question 5 9 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2012
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Double angle with reciprocal functions
Difficulty	Standard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) requires applying double angle formula and reciprocal definitions (routine manipulation), part (b) follows directly from (a) using basic identities, and part (c) uses the result from (b) to solve an equation. While it involves reciprocal trig functions which students find less familiar, the step-by-step guidance and standard techniques make it slightly easier than average for C3.
Spec	1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 1.05o Trigonometric equations: solve in given intervals 1.05p Proof involving trig: functions and identities

(a) Express $4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta$ in terms of $\sin \theta$ and $\cos \theta$.
(b) Hence show that

$$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta$$ (c) Hence or otherwise solve, for $0 < \theta < \pi$, $$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = 4$$ giving your answers in terms of $\pi$.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $4\cos\sec^2 2\theta - \cos\sec^2 \theta = \frac{4}{\sin^2 2\theta} - \frac{1}{\sin^2 \theta}$	B1	One term correct. Eg writes $4\cos\sec^2 2\theta$ as $\frac{4}{(2\sin\theta\cos\theta)^2}$ or $\cos\sec^2 \theta$ as $\frac{1}{\sin^2 \theta}$. Accept terms like $\cos\sec^2 \theta = 1 + \cot^2 \theta = 1 + \frac{\cos^2\theta}{\sin^2\theta}$. The question merely asks for an expression in $\sin\theta$ and $\cos\theta$
$= \frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$	B1	A fully correct expression in $\sin\theta$ and $\cos\theta$. Eg. $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$. Accept equivalents. Allow a different variable say $x$'s instead of $\theta$'s but do not allow mixed units.
(2)	(2)
(b) $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta} = \frac{4}{\sin^2\theta} - \frac{4}{A\sin^2\theta\cos^2\theta} - \frac{1}{\sin^2\theta}$	M1	Attempts to combine their expression in $\sin\theta$ and $\cos\theta$ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted.
$= \frac{1}{\sin^2\theta\cos^2\theta} - \frac{\cos^2\theta}{\sin^2\theta\cos^2\theta}$	M1	Attempts to form a 'single' term on the numerator by using the identity $1 - \cos^2 \theta = \sin^2 \theta$
$= \frac{\sin^2\theta}{\sin^2\theta\cos^2\theta}$	M1	Cancels correctly by $\sin^2\theta$ terms and replaces $\frac{1}{\cos^2\theta}$ with $\sec^2\theta$
$= \frac{1}{\cos^2\theta} = \sec^2\theta$	M1A1*	Cso. All aspects must be correct
(4)	(4)
(c) $\sec^2\theta = 4 \Rightarrow \sec\theta = \pm 2 \Rightarrow \cos\theta = \pm \frac{1}{2}$	M1	For $\sec^2\theta = 4$ leading to a solution of $\cos\theta$ by taking the root and inverting in either order. Similarly accept $\tan^2\theta = 3, \sin^2\theta = \frac{3}{4}$ leading to solutions of $\tan\theta, \sin\theta$. Also accept $\cos 2\theta = -\frac{1}{2}$
$\theta = \frac{\pi}{3}, \frac{2\pi}{3}$	A1, A1	Obtains one correct answer usually $\theta = \frac{\pi}{3}$. Do not accept decimal answers or degrees.
(3)	(9 marks)

Note (a) and (b) can be scored together.

Notes for (a) and (b):

- (a) B1: One term correct. Eg writes $4\cos\sec^2 2\theta$ as $\frac{4}{(2\sin\theta\cos\theta)^2}$ or $\cos\sec^2 \theta$ as $\frac{1}{\sin^2\theta}$. Accept terms like $\cos\sec^2 \theta = 1 + \cot^2 \theta = 1 + \frac{\cos^2\theta}{\sin^2\theta}$. The question merely asks for an expression in $\sin\theta$ and $\cos\theta$.

- B1: A fully correct expression in $\sin\theta$ and $\cos\theta$. Eg. $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$. Accept equivalents. Allow a different variable say $x$'s instead of $\theta$'s but do not allow mixed units.

- b) M1: Attempts to combine their expression in $\sin\theta$ and $\cos\theta$ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted.

- M1: Attempts to form a 'single' term on the numerator by using the identity $1 - \cos^2 \theta = \sin^2 \theta$.

- M1: Cancels correctly by $\sin^2\theta$ terms and replaces $\frac{1}{\cos^2\theta}$ with $\sec^2\theta$.

- M1A1*: Cso. All aspects must be correct.

IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER

| (a) $4\cos\sec^2 2\theta - \cos\sec^2 \theta = \frac{4}{\sin^2 2\theta} - \frac{1}{\sin^2 \theta}$ | B1 | One term correct. Eg writes $4\cos\sec^2 2\theta$ as $\frac{4}{(2\sin\theta\cos\theta)^2}$ or $\cos\sec^2 \theta$ as $\frac{1}{\sin^2 \theta}$. Accept terms like $\cos\sec^2 \theta = 1 + \cot^2 \theta = 1 + \frac{\cos^2\theta}{\sin^2\theta}$. The question merely asks for an expression in $\sin\theta$ and $\cos\theta$ |
|---|---|---|
| $= \frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$ | B1 | A fully correct expression in $\sin\theta$ and $\cos\theta$. Eg. $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$. Accept equivalents. Allow a different variable say $x$'s instead of $\theta$'s but do not allow mixed units. |
| **(2)** | **(2)** | |
| (b) $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta} = \frac{4}{\sin^2\theta} - \frac{4}{A\sin^2\theta\cos^2\theta} - \frac{1}{\sin^2\theta}$ | M1 | Attempts to combine their expression in $\sin\theta$ and $\cos\theta$ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted. |
| $= \frac{1}{\sin^2\theta\cos^2\theta} - \frac{\cos^2\theta}{\sin^2\theta\cos^2\theta}$ | M1 | Attempts to form a 'single' term on the numerator by using the identity $1 - \cos^2 \theta = \sin^2 \theta$ |
| $= \frac{\sin^2\theta}{\sin^2\theta\cos^2\theta}$ | M1 | Cancels correctly by $\sin^2\theta$ terms and replaces $\frac{1}{\cos^2\theta}$ with $\sec^2\theta$ |
| $= \frac{1}{\cos^2\theta} = \sec^2\theta$ | M1A1* | Cso. All aspects must be correct |
| **(4)** | **(4)** | |
| (c) $\sec^2\theta = 4 \Rightarrow \sec\theta = \pm 2 \Rightarrow \cos\theta = \pm \frac{1}{2}$ | M1 | For $\sec^2\theta = 4$ leading to a solution of $\cos\theta$ by taking the root and inverting in either order. Similarly accept $\tan^2\theta = 3, \sin^2\theta = \frac{3}{4}$ leading to solutions of $\tan\theta, \sin\theta$. Also accept $\cos 2\theta = -\frac{1}{2}$ |
| $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$ | A1, A1 | Obtains one correct answer usually $\theta = \frac{\pi}{3}$. Do not accept decimal answers or degrees. |
| **(3)** | **(9 marks)** | |

**Note (a) and (b) can be scored together.**

**Notes for (a) and (b):**
- (a) B1: One term correct. Eg writes $4\cos\sec^2 2\theta$ as $\frac{4}{(2\sin\theta\cos\theta)^2}$ or $\cos\sec^2 \theta$ as $\frac{1}{\sin^2\theta}$. Accept terms like $\cos\sec^2 \theta = 1 + \cot^2 \theta = 1 + \frac{\cos^2\theta}{\sin^2\theta}$. The question merely asks for an expression in $\sin\theta$ and $\cos\theta$.
- B1: A fully correct expression in $\sin\theta$ and $\cos\theta$. Eg. $\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}$. Accept equivalents. Allow a different variable say $x$'s instead of $\theta$'s but do not allow mixed units.
- b) M1: Attempts to combine their expression in $\sin\theta$ and $\cos\theta$ using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted.
- M1: Attempts to form a 'single' term on the numerator by using the identity $1 - \cos^2 \theta = \sin^2 \theta$.
- M1: Cancels correctly by $\sin^2\theta$ terms and replaces $\frac{1}{\cos^2\theta}$ with $\sec^2\theta$.
- M1A1*: Cso. All aspects must be correct.

**IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER**

---

Show LaTeX source

\begin{enumerate}
  \item (a) Express $4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta$ in terms of $\sin \theta$ and $\cos \theta$.\\
(b) Hence show that
\end{enumerate}

$$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = \sec ^ { 2 } \theta$$

(c) Hence or otherwise solve, for $0 < \theta < \pi$,

$$4 \operatorname { cosec } ^ { 2 } 2 \theta - \operatorname { cosec } ^ { 2 } \theta = 4$$

giving your answers in terms of $\pi$.\\

\hfill \mbox{\textit{Edexcel C3 2012 Q5 [9]}}

This paper (8 questions)

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Q1 4 Q2 9 Q3 9 Q4 7 Q5 9 Q6 14 Q7 11 Q8 12

Edexcel C3 2012 June — Question 5 9 marks

Note (a) and (b) can be scored together.

Notes for (a) and (b):

- B1: A fully correct expression in \(\sin\theta\) and \(\cos\theta\). Eg. \(\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}\). Accept equivalents. Allow a different variable say \(x\)'s instead of \(\theta\)'s but do not allow mixed units.

- b) M1: Attempts to combine their expression in \(\sin\theta\) and \(\cos\theta\) using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted.

- M1: Attempts to form a 'single' term on the numerator by using the identity \(1 - \cos^2 \theta = \sin^2 \theta\).

- M1: Cancels correctly by \(\sin^2\theta\) terms and replaces \(\frac{1}{\cos^2\theta}\) with \(\sec^2\theta\).

- M1A1*: Cso. All aspects must be correct.

IF IN ANY DOUBT SEND TO REVIEW OR CONSULT YOUR TEAM LEADER

This paper (8 questions)

Answer	Marks	Guidance
(a) \(4\cos\sec^2 2\theta - \cos\sec^2 \theta = \frac{4}{\sin^2 2\theta} - \frac{1}{\sin^2 \theta}\)	B1	One term correct. Eg writes \(4\cos\sec^2 2\theta\) as \(\frac{4}{(2\sin\theta\cos\theta)^2}\) or \(\cos\sec^2 \theta\) as \(\frac{1}{\sin^2 \theta}\). Accept terms like \(\cos\sec^2 \theta = 1 + \cot^2 \theta = 1 + \frac{\cos^2\theta}{\sin^2\theta}\). The question merely asks for an expression in \(\sin\theta\) and \(\cos\theta\)
\(= \frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}\)	B1	A fully correct expression in \(\sin\theta\) and \(\cos\theta\). Eg. \(\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta}\). Accept equivalents. Allow a different variable say \(x\)'s instead of \(\theta\)'s but do not allow mixed units.
(2)	(2)
(b) \(\frac{4}{(2\sin\theta\cos\theta)^2} - \frac{1}{\sin^2\theta} = \frac{4}{\sin^2\theta} - \frac{4}{A\sin^2\theta\cos^2\theta} - \frac{1}{\sin^2\theta}\)	M1	Attempts to combine their expression in \(\sin\theta\) and \(\cos\theta\) using a common denominator. The terms can be separate but the denominator must be correct and one of the numerators must have been adapted.
\(= \frac{1}{\sin^2\theta\cos^2\theta} - \frac{\cos^2\theta}{\sin^2\theta\cos^2\theta}\)	M1	Attempts to form a 'single' term on the numerator by using the identity \(1 - \cos^2 \theta = \sin^2 \theta\)
\(= \frac{\sin^2\theta}{\sin^2\theta\cos^2\theta}\)	M1	Cancels correctly by \(\sin^2\theta\) terms and replaces \(\frac{1}{\cos^2\theta}\) with \(\sec^2\theta\)
\(= \frac{1}{\cos^2\theta} = \sec^2\theta\)	M1A1*	Cso. All aspects must be correct
(4)	(4)
(c) \(\sec^2\theta = 4 \Rightarrow \sec\theta = \pm 2 \Rightarrow \cos\theta = \pm \frac{1}{2}\)	M1	For \(\sec^2\theta = 4\) leading to a solution of \(\cos\theta\) by taking the root and inverting in either order. Similarly accept \(\tan^2\theta = 3, \sin^2\theta = \frac{3}{4}\) leading to solutions of \(\tan\theta, \sin\theta\). Also accept \(\cos 2\theta = -\frac{1}{2}\)
\(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)	A1, A1	Obtains one correct answer usually \(\theta = \frac{\pi}{3}\). Do not accept decimal answers or degrees.
(3)	(9 marks)