| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Multiple independent equations — includes show/prove component |
| Difficulty | Standard +0.3 This is a standard C4 trigonometric equations question requiring double angle formulas, factorization, and quadratic formula application. Part (a) uses tan 2x expansion and basic algebra; part (b) applies sin 2x and cos 2x identities followed by routine quadratic solving. All techniques are core syllabus with clear signposting, making it slightly easier than average. |
| 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 | Mark | Guidance |
| \(\tan 2x = \frac{2\tan x}{1-\tan^2 x}\) substituted | M1 | Use of double angle formula |
| \(\frac{2\tan x}{1-\tan^2 x} + \tan x = 0\) | ||
| \(2\tan x + \tan x(1-\tan^2 x) = 0\) | M1 | Multiply through by \((1-\tan^2 x)\) |
| \(\tan x(3 - \tan^2 x) = 0\) | A1 | Hence \(\tan x = 0\) or \(\tan^2 x = 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 0°, 60°, 120°, 180°\) | B1 | All four values, no extras in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin 2x = \cos x \cos 2x\), divide by \(\cos x\): \(2\sin x = \cos 2x\) | M1 | Use \(\sin 2x = 2\sin x \cos x\) and divide by \(\cos x\) |
| \(2\sin x = 1 - 2\sin^2 x\) | M1 | Use \(\cos 2x = 1-2\sin^2 x\) |
| \(2\sin^2 x + 2\sin x - 1 = 0\) | A1 | Correct rearrangement shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin x = \frac{-2 \pm \sqrt{4+8}}{4} = \frac{-2 \pm \sqrt{12}}{4}\) | M1 | Apply quadratic formula |
| \(= \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}\) | A1 | Simplify |
| Reject \(\frac{-1-\sqrt{3}}{2}\) as \(< -1\) | M1 | Valid reason for rejection |
| \(\sin x = \frac{\sqrt{3}-1}{2}\), so \(p = 2\) | A1 | Correct value identified |
## Question 6:
**Part (a)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan 2x = \frac{2\tan x}{1-\tan^2 x}$ substituted | M1 | Use of double angle formula |
| $\frac{2\tan x}{1-\tan^2 x} + \tan x = 0$ | | |
| $2\tan x + \tan x(1-\tan^2 x) = 0$ | M1 | Multiply through by $(1-\tan^2 x)$ |
| $\tan x(3 - \tan^2 x) = 0$ | A1 | Hence $\tan x = 0$ or $\tan^2 x = 3$ |
**Part (a)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0°, 60°, 120°, 180°$ | B1 | All four values, no extras in range |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin 2x = \cos x \cos 2x$, divide by $\cos x$: $2\sin x = \cos 2x$ | M1 | Use $\sin 2x = 2\sin x \cos x$ and divide by $\cos x$ |
| $2\sin x = 1 - 2\sin^2 x$ | M1 | Use $\cos 2x = 1-2\sin^2 x$ |
| $2\sin^2 x + 2\sin x - 1 = 0$ | A1 | Correct rearrangement shown |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin x = \frac{-2 \pm \sqrt{4+8}}{4} = \frac{-2 \pm \sqrt{12}}{4}$ | M1 | Apply quadratic formula |
| $= \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$ | A1 | Simplify |
| Reject $\frac{-1-\sqrt{3}}{2}$ as $< -1$ | M1 | Valid reason for rejection |
| $\sin x = \frac{\sqrt{3}-1}{2}$, so $p = 2$ | A1 | Correct value identified |
---6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\tan 2 x + \tan x = 0$, show that $\tan x = 0$ or $\tan ^ { 2 } x = 3$.
\item Hence find all solutions of $\tan 2 x + \tan x = 0$ in the interval $0 ^ { \circ } < x < 180 ^ { \circ }$.\\
(l mark)
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $\cos x \neq 0$, show that the equation
$$\sin 2 x = \cos x \cos 2 x$$
can be written in the form
$$2 \sin ^ { 2 } x + 2 \sin x - 1 = 0$$
\item Show that all solutions of the equation $2 \sin ^ { 2 } x + 2 \sin x - 1 = 0$ are given by $\sin x = \frac { \sqrt { 3 } - 1 } { p }$, where $p$ is an integer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2011 Q6 [10]}}