| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.8 This question requires proving a non-trivial trigonometric identity involving double angles and reciprocal functions, then applying it to solve equations including one with compound angles. Part (a) demands algebraic manipulation across multiple trig identities (double angle formulas, tan = sin/cos). Part (b)(ii) is particularly challenging as students must recognize the structure matches part (a) after substitution. The multi-step reasoning and need to connect the proof to the applications elevates this above standard C3/C4 fare, though it remains within typical A-level scope. |
| 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 |
|---|---|---|
| \(\sin 2\theta - \tan\theta = \sqrt{3}\cos 2\theta \Rightarrow \tan\theta\cos 2\theta = \sqrt{3}\cos 2\theta\) | Opening simplification shown | |
| \(\tan\theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}\) (awrt 1.05) | M1A1 | M1: \(\tan\theta = \pm\sqrt{3} \Rightarrow \theta = \ldots\) A1: \(\theta = \frac{\pi}{3}\), accept awrt 1.05. Ignore solutions outside range but withhold A mark for extra solutions in range. |
| \(\cos 2\theta = 0 \Rightarrow \theta = \frac{\pi}{4}\) (awrt 0.785) | M1A1 | M1: \(\cos 2\theta = 0 \Rightarrow \theta = \ldots\) A1: \(\theta = \frac{\pi}{4}\), accept awrt 0.785. Ignore solutions outside range but withhold A mark for extra solutions in range. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan(\theta+1)\cos(2\theta+2) - \sin(2\theta+2) = 2 \Rightarrow \tan(\theta+1) = -2\) | M1 | M1: \(\tan(\theta+1) = \pm 2\) |
| \(\Rightarrow \theta = \arctan(-2) - 1\) | dM1 | Correct order of operations i.e. \(\theta = \arctan(\pm 2) - 1\). May be implied by \(\theta = -2.1\ldots\) |
| \(\Rightarrow \theta = 1.03\) | A1 | awrt \(\theta = 1.03\). Ignore solutions outside range but withhold A mark for extra solutions in range. |
## Question 8(b)(i):
$\sin 2\theta - \tan\theta = \sqrt{3}\cos 2\theta \Rightarrow \tan\theta\cos 2\theta = \sqrt{3}\cos 2\theta$ | | Opening simplification shown
$\tan\theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3}$ (awrt 1.05) | M1A1 | M1: $\tan\theta = \pm\sqrt{3} \Rightarrow \theta = \ldots$ A1: $\theta = \frac{\pi}{3}$, accept awrt 1.05. Ignore solutions outside range but withhold A mark for extra solutions in range.
$\cos 2\theta = 0 \Rightarrow \theta = \frac{\pi}{4}$ (awrt 0.785) | M1A1 | M1: $\cos 2\theta = 0 \Rightarrow \theta = \ldots$ A1: $\theta = \frac{\pi}{4}$, accept awrt 0.785. Ignore solutions outside range but withhold A mark for extra solutions in range.
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## Question 8(b)(ii):
$\tan(\theta+1)\cos(2\theta+2) - \sin(2\theta+2) = 2 \Rightarrow \tan(\theta+1) = -2$ | M1 | M1: $\tan(\theta+1) = \pm 2$
$\Rightarrow \theta = \arctan(-2) - 1$ | dM1 | Correct order of operations i.e. $\theta = \arctan(\pm 2) - 1$. May be implied by $\theta = -2.1\ldots$
$\Rightarrow \theta = 1.03$ | A1 | awrt $\theta = 1.03$. Ignore solutions outside range but withhold A mark for extra solutions in range.
---8. (a) Prove that
$$\sin 2 x - \tan x \equiv \tan x \cos 2 x , \quad x \neq \frac { ( 2 n + 1 ) \pi } { 2 } , \quad n \in \mathbb { Z }$$
(b) Hence solve, for $0 \leqslant \theta < \frac { \pi } { 2 }$
\begin{enumerate}[label=(\roman*)]
\item $\sin 2 \theta - \tan \theta = \sqrt { 3 } \cos 2 \theta$
\item $\tan ( \theta + 1 ) \cos ( 2 \theta + 2 ) - \sin ( 2 \theta + 2 ) = 2$
Give your answers in radians to 3 significant figures, as appropriate.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2016 Q8 [11]}}