| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Compound angle with reciprocal functions |
| Difficulty | Standard +0.8 This question requires proving a non-standard identity involving reciprocal trig functions and double angles, then applying it to solve an equation. Part (a) demands algebraic manipulation of sec and tan with double angle formulas—not a routine textbook identity. Part (b) requires substitution, algebraic rearrangement, and solving a resulting trigonometric equation over a full period. The combination of reciprocal functions, compound angles, and multi-step problem-solving places this above average difficulty but not at the extreme end. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec 2A = \frac{1}{\cos 2A}\) OR \(\tan 2A = \frac{\sin 2A}{\cos 2A}\) | B1 | Need not be in proof; implied by sight of \(\frac{1}{\cos^2 A - \sin^2 A}\) |
| \(\sec 2A + \tan 2A = \frac{1+\sin 2A}{\cos 2A}\) | M1 | Setting as single fraction; denominator correct, at least two terms on numerator |
| \(= \frac{1+2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | M1 | Getting expression in just \(\sin A\) and \(\cos A\) using double angle identities |
| \(= \frac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A}\) | M1 | Replacing 1 by \(\cos^2 A + \sin^2 A\) and factorising both numerator and denominator |
| \(= \frac{(\cos A + \sin A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)} = \frac{\cos A + \sin A}{\cos A - \sin A}\) | A1* | Cancelling to produce given answer with no errors; mixing variables loses A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sec 2\theta + \tan 2\theta = \frac{1}{2} \Rightarrow \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1}{2}\) | Using part (a) | |
| \(2\cos\theta + 2\sin\theta = \cos\theta - \sin\theta\) | M1 | Cross multiplying, dividing by \(\cos\theta\) to reach \(\tan\theta = k\); condone \(\tan 2\theta = k\) for this mark only |
| \(\tan\theta = -\frac{1}{3}\) | A1 | |
| \(\theta =\) awrt \(2.820, 5.961\) | dM1 A1 | Dependent on \(\tan\theta = k\) leading to at least one value (1 dp); no extra solutions in range; condone 2.82 for 2.820; allow answers in degrees for dM1 |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec 2A = \frac{1}{\cos 2A}$ **OR** $\tan 2A = \frac{\sin 2A}{\cos 2A}$ | B1 | Need not be in proof; implied by sight of $\frac{1}{\cos^2 A - \sin^2 A}$ |
| $\sec 2A + \tan 2A = \frac{1+\sin 2A}{\cos 2A}$ | M1 | Setting as single fraction; denominator correct, at least two terms on numerator |
| $= \frac{1+2\sin A\cos A}{\cos^2 A - \sin^2 A}$ | M1 | Getting expression in just $\sin A$ and $\cos A$ using double angle identities |
| $= \frac{\cos^2 A + \sin^2 A + 2\sin A\cos A}{\cos^2 A - \sin^2 A}$ | M1 | Replacing 1 by $\cos^2 A + \sin^2 A$ **and** factorising both numerator and denominator |
| $= \frac{(\cos A + \sin A)(\cos A + \sin A)}{(\cos A + \sin A)(\cos A - \sin A)} = \frac{\cos A + \sin A}{\cos A - \sin A}$ | A1* | Cancelling to produce given answer with no errors; mixing variables loses A1* |
**(5 marks)**
---
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sec 2\theta + \tan 2\theta = \frac{1}{2} \Rightarrow \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} = \frac{1}{2}$ | | Using part (a) |
| $2\cos\theta + 2\sin\theta = \cos\theta - \sin\theta$ | M1 | Cross multiplying, dividing by $\cos\theta$ to reach $\tan\theta = k$; condone $\tan 2\theta = k$ for this mark only |
| $\tan\theta = -\frac{1}{3}$ | A1 | |
| $\theta =$ awrt $2.820, 5.961$ | dM1 A1 | Dependent on $\tan\theta = k$ leading to at least one value (1 dp); no extra solutions in range; condone 2.82 for 2.820; allow answers in degrees for dM1 |
**(4 marks)**
---\begin{enumerate}
\item (a) Prove that
\end{enumerate}
$$\sec 2 A + \tan 2 A \equiv \frac { \cos A + \sin A } { \cos A - \sin A } , \quad A \neq \frac { ( 2 n + 1 ) \pi } { 4 } , n \in \mathbb { Z }$$
(b) Hence solve, for $0 \leqslant \theta < 2 \pi$,
$$\sec 2 \theta + \tan 2 \theta = \frac { 1 } { 2 }$$
Give your answers to 3 decimal places.\\
\hfill \mbox{\textit{Edexcel C3 2015 Q8 [9]}}