| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| 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 requiring standard double angle identities and algebraic manipulation. Part (a) involves routine identity verification using cos 2x = cos²x - sin²x and factorization. Part (b) applies the results from (a) with straightforward substitution. Part (c) is a standard equation solving tan 2θ = 1. While it requires multiple techniques, each step follows predictable patterns with no novel insight needed, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05j 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(\cos 2x = \cos^2 x - \sin^2 x\) in attempt to prove identity | M1 | |
| \(\frac{\cos 2x}{\cos x + \sin x} = \frac{\cos^2 x - \sin^2 x}{\cos x + \sin x} = \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x + \sin x} = \cos x - \sin x\) * | A1 | cso (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(\cos 2x = 2\cos^2 x - 1\) in attempt to prove identity | M1 | |
| Use of \(\sin 2x = 2\sin x\cos x\) in attempt to prove identity | M1 | |
| \(\frac{1}{2}(\cos 2x - \sin 2x) = \frac{1}{2}(2\cos^2 x - 1 - 2\sin x\cos x) = \cos^2 x - \cos x\sin x - \frac{1}{2}\) * | A1 | cso (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta(\cos\theta - \sin\theta) = \frac{1}{2}\) using (a)(i) | M1 | |
| \(\cos^2\theta - \cos\theta\sin\theta - \frac{1}{2} = 0\) | ||
| \(\frac{1}{2}(\cos 2\theta - \sin 2\theta) = 0\) using (a)(ii) | M1 | |
| \(\cos 2\theta = \sin 2\theta\) * | A1 | cso (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan 2\theta = 1\) | M1 | |
| \(2\theta = \frac{\pi}{4}, \left(\frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}\right)\) | A1 | any one correct value of \(2\theta\) |
| \(\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}\) | M1 | Obtaining at least 2 solutions in range |
| All 4 correct solutions | A1 | (4) [12] |
# Question 7:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $\cos 2x = \cos^2 x - \sin^2 x$ in attempt to prove identity | M1 | |
| $\frac{\cos 2x}{\cos x + \sin x} = \frac{\cos^2 x - \sin^2 x}{\cos x + \sin x} = \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x + \sin x} = \cos x - \sin x$ * | A1 | cso **(2)** |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $\cos 2x = 2\cos^2 x - 1$ in attempt to prove identity | M1 | |
| Use of $\sin 2x = 2\sin x\cos x$ in attempt to prove identity | M1 | |
| $\frac{1}{2}(\cos 2x - \sin 2x) = \frac{1}{2}(2\cos^2 x - 1 - 2\sin x\cos x) = \cos^2 x - \cos x\sin x - \frac{1}{2}$ * | A1 | cso **(3)** |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta(\cos\theta - \sin\theta) = \frac{1}{2}$ using (a)(i) | M1 | |
| $\cos^2\theta - \cos\theta\sin\theta - \frac{1}{2} = 0$ | | |
| $\frac{1}{2}(\cos 2\theta - \sin 2\theta) = 0$ using (a)(ii) | M1 | |
| $\cos 2\theta = \sin 2\theta$ * | A1 | cso **(3)** |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 2\theta = 1$ | M1 | |
| $2\theta = \frac{\pi}{4}, \left(\frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}\right)$ | A1 | any one correct value of $2\theta$ |
| $\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$ | M1 | Obtaining at least 2 solutions in range |
| All 4 correct solutions | A1 | **(4) [12]** |7. (a) Show that
\begin{enumerate}[label=(\roman*)]
\item $\frac { \cos 2 x } { \cos x + \sin x } \equiv \cos x - \sin x , \quad x \neq \left( n - \frac { 1 } { 4 } \right) \pi , n \in \mathbb { Z }$,
\item $\frac { 1 } { 2 } ( \cos 2 x - \sin 2 x ) \equiv \cos ^ { 2 } x - \cos x \sin x - \frac { 1 } { 2 }$.\\
(b) Hence, or otherwise, show that the equation
$$\cos \theta \left( \frac { \cos 2 \theta } { \cos \theta + \sin \theta } \right) = \frac { 1 } { 2 }$$
can be written as
$$\sin 2 \theta = \cos 2 \theta$$
(c) Solve, for $0 \leqslant \theta < 2 \pi$,
$$\sin 2 \theta = \cos 2 \theta$$
giving your answers in terms of $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2006 Q7 [12]}}