| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Moderate -0.3 Part (a) is a standard double-angle identity proof requiring straightforward application of sin 2θ = 2sinθcosθ and 1 + cos 2θ = 2cos²θ. Part (b) follows directly from (a), reducing to tan θ = 1/2, then solving in the given range. This is routine C3 material with no novel insight required, slightly easier than average due to the 'hence' structure guiding students through. |
| Spec | 1.05j 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 |
| \(\frac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1}\) | M1 | Uses both a correct identity for \(\sin 2\theta\) and a correct identity for \(\cos 2\theta\). Also allow \(1 + \cos 2\theta = 2\cos^2\theta\) on denominator. Angles must be consistent. |
| \(\frac{\cancel{2}\sin\theta\cancel{\cos\theta}}{\cancel{2}\cos\theta\cancel{\cos\theta}} = \tan\theta\) AG | A1 cso | Correct proof. No errors seen. |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\tan\theta = 1 \Rightarrow \tan\theta = \frac{1}{2}\) | M1 | For either \(2\tan\theta = 1\) or \(\tan\theta = \frac{1}{2}\), seen or implied. |
| \(\theta_1 = \text{awrt } 26.6°\) | A1 | awrt 26.6 |
| \(\theta_2 = \text{awrt } -153.4°\) | A1\(\checkmark\) | awrt \(-153.4°\) or \(\theta_2 = -180° + \theta_1\). Special Case: \(\tan\theta = k\), \(k \neq \frac{1}{2}\), giving \(\theta_2 = -180° + \theta_1\): award M0A0B1. Candidates writing \(\tan\theta = 1\) giving only \(45°\) and \(-135°\): SC M0A0B1. |
| (3) [5] |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1}$ | M1 | Uses **both** a correct identity for $\sin 2\theta$ **and** a correct identity for $\cos 2\theta$. Also allow $1 + \cos 2\theta = 2\cos^2\theta$ on denominator. Angles must be consistent. |
| $\frac{\cancel{2}\sin\theta\cancel{\cos\theta}}{\cancel{2}\cos\theta\cancel{\cos\theta}} = \tan\theta$ **AG** | A1 cso | Correct proof. No errors seen. |
| | **(2)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\tan\theta = 1 \Rightarrow \tan\theta = \frac{1}{2}$ | M1 | For either $2\tan\theta = 1$ or $\tan\theta = \frac{1}{2}$, seen or implied. |
| $\theta_1 = \text{awrt } 26.6°$ | A1 | awrt 26.6 |
| $\theta_2 = \text{awrt } -153.4°$ | A1$\checkmark$ | awrt $-153.4°$ or $\theta_2 = -180° + \theta_1$. **Special Case:** $\tan\theta = k$, $k \neq \frac{1}{2}$, giving $\theta_2 = -180° + \theta_1$: award M0A0B1. Candidates writing $\tan\theta = 1$ giving only $45°$ and $-135°$: SC M0A0B1. |
| | **(3) [5]** | |
---\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$\frac { \sin 2 \theta } { 1 + \cos 2 \theta } = \tan \theta$$
(b) Hence find, for $- 180 ^ { \circ } \leqslant \theta < 180 ^ { \circ }$, all the solutions of
$$\frac { 2 \sin 2 \theta } { 1 + \cos 2 \theta } = 1$$
Give your answers to 1 decimal place.\\
\hfill \mbox{\textit{Edexcel C3 2010 Q1 [5]}}