| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | General solution — then find specific solutions |
| Difficulty | Moderate -0.3 Part (a) is trivial recall of complementary angles. Part (b) requires converting sin to cos using the identity and solving a standard trigonometric equation—routine for FP1. Part (c) asks to find a unique tan value from the general solution, requiring some algebraic manipulation but following directly from (b). Overall slightly easier than average due to the guided structure and standard techniques, though the final part adds minor complexity. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(k = 6\) | B1 | A correct value of \(k\). Either \(6\) or \(-6\) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(\cos\left(2x - \frac{5\pi}{6}\right) = \cos\frac{\pi}{6}\) | M1 | \(\cos\left(2x - \frac{5\pi}{6}\right) = \cos\frac{\pi}{k}\) stated or used; if incorrect ft c's \(k\) value |
| Altn: \(\sin\left[\frac{\pi}{2} - \left(2x - \frac{5\pi}{6}\right)\right] = \sin\frac{\pi}{3}\) (*) | OE | |
| \(2x - \frac{5\pi}{6} = 2n\pi + \frac{\pi}{6}\) | m1 | OE Either one, showing a correct use of \(2n\pi\) in forming a general solution. If incorrect ft c's \(k\) value |
| \(2x - \frac{5\pi}{6} = 2n\pi - \frac{\pi}{6}\) | ||
| Altn: using (*) above, | OE \(X = 2n\pi + \frac{\pi}{3}\) or \(X = 2n\pi + \pi - \frac{\pi}{3}\) OE where \(X = \frac{\pi}{2} - \left(2x - \frac{5\pi}{6}\right)\) | |
| Condone \(360n\) for \(2n\pi\) in both methods | ||
| \(x = \frac{1}{2}\left(2n\pi + \frac{5\pi}{6} + \frac{\pi}{6}\right)\) | A1F | Full sets of GS, if incorrect ft c's \(k\) value, condoning unsimplified forms ie check \(x = \frac{1}{2}\left(2n\pi + \frac{5\pi}{6} \pm \frac{\pi}{k}\right)\) |
| (A0F if degrees present in answer) | ||
| \(x = n\pi + \frac{\pi}{2}\), \(x = n\pi + \frac{\pi}{3}\) | A1 | OE full set of correct solutions in rads with constant terms combined |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \(\tan x = \tan(n\pi + \frac{\pi}{2}) = \tan\frac{\pi}{2}\) not finite | E1 | Considers the complete set of general solutions from (b), showing that one results in non finite value for tan \(x\) and the other gives single value. Must be working with general \(n\) and must refer to either 'finite' or 'non finite' |
| \(\tan x = \tan(n\pi + \frac{\pi}{3}) = \tan\frac{\pi}{3}\) (ie a single finite value) | ||
| (Only possible finite value for tan \(x\) is) \(\sqrt{3}\) | B1 | \(\sqrt{3}\) This B1 mark is dep on \(k=6\) and at least 3 marks scored in part (b) but not dependent on E1 |
**(a)** $k = 6$ | B1 | A correct value of $k$. Either $6$ or $-6$
**Total for (a): 1 mark**
**(b)** $\cos\left(2x - \frac{5\pi}{6}\right) = \cos\frac{\pi}{6}$ | M1 | $\cos\left(2x - \frac{5\pi}{6}\right) = \cos\frac{\pi}{k}$ stated or used; if incorrect ft c's $k$ value
**Altn:** $\sin\left[\frac{\pi}{2} - \left(2x - \frac{5\pi}{6}\right)\right] = \sin\frac{\pi}{3}$ (*) | | OE
$2x - \frac{5\pi}{6} = 2n\pi + \frac{\pi}{6}$ | m1 | OE Either one, showing a correct use of $2n\pi$ in forming a general solution. If incorrect ft c's $k$ value
$2x - \frac{5\pi}{6} = 2n\pi - \frac{\pi}{6}$ | |
**Altn:** using (*) above, | | OE $X = 2n\pi + \frac{\pi}{3}$ or $X = 2n\pi + \pi - \frac{\pi}{3}$ OE where $X = \frac{\pi}{2} - \left(2x - \frac{5\pi}{6}\right)$
| | Condone $360n$ for $2n\pi$ in both methods
$x = \frac{1}{2}\left(2n\pi + \frac{5\pi}{6} + \frac{\pi}{6}\right)$ | A1F | Full sets of GS, if incorrect ft c's $k$ value, condoning unsimplified forms ie check $x = \frac{1}{2}\left(2n\pi + \frac{5\pi}{6} \pm \frac{\pi}{k}\right)$
| | **(A0F if degrees present in answer)**
$x = n\pi + \frac{\pi}{2}$, $x = n\pi + \frac{\pi}{3}$ | A1 | OE full set of correct solutions in rads with constant terms combined
**Total for (b): 4 marks**
**(c)** $\tan x = \tan(n\pi + \frac{\pi}{2}) = \tan\frac{\pi}{2}$ not finite | E1 | Considers the complete set of general solutions from (b), showing that one results in non finite value for tan $x$ and the other gives single value. Must be working with general $n$ and must refer to either 'finite' or 'non finite'
$\tan x = \tan(n\pi + \frac{\pi}{3}) = \tan\frac{\pi}{3}$ (ie a single finite value) | |
(Only possible finite value for tan $x$ is) $\sqrt{3}$ | B1 | $\sqrt{3}$ This B1 mark is dep on $k=6$ and at least 3 marks scored in part (b) but not dependent on E1
**Total for (c): 2 marks**
**Overall Total: 7 marks**
**Additional Notes:**
- (b) Example: $\cos\left(2x - \frac{5\pi}{6}\right) = \cos\frac{\pi}{6}$, $2x - \frac{5\pi}{6} = \frac{\pi}{6}$, $2x = \pi$, $2x = 2n\pi \pm \pi$, $x = n\pi \pm \frac{\pi}{2}$ (M1m0)
---\begin{enumerate}[label=(\alph*)]
\item Given that $\sin \frac{\pi}{3} = \cos \frac{\pi}{k}$, state the value of the integer $k$.
[1 mark]
\item Hence, or otherwise, find the general solution of the equation
$$\cos \left( 2x - \frac{5\pi}{6} \right) = \sin \frac{\pi}{3}$$
giving your answer, in its simplest form, in terms of $\pi$.
[4 marks]
\item Hence, given that $\cos \left( 2x - \frac{5\pi}{6} \right) = \sin \frac{\pi}{3}$, show that there is only one finite value for $\tan x$ and state its exact value.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2016 Q4 [7]}}