| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve factored trig equation |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation question with routine algebraic manipulation. Part (a) involves basic manipulation to find tan x and solving a factored equation. Part (b) requires substituting sin²θ = 1 - cos²θ to get a quadratic in cos θ, then completing the square or using calculus to find the minimum—all standard techniques for this level with no novel insight required. |
| 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)(i) | \(\frac{4\sin x}{\cos x} + \frac{5\cos x}{\cos x} = 0; \quad 4\tan x + 5 = 0\) | M1 |
| (a)(i) | \(\tan x = -\frac{5}{4}\) \((= -1.25)\) | A1 |
| (a)(ii) | \(\tan x = 1, \quad \tan x = -1.25\) | B1F |
| (a)(ii) | \((x =) 45°, 225°, 129°, 309°\) | B2, 1 |
| Total | 9 | |
| (b) | \(\frac{16 + 9\sin^2 \theta}{5 - 3\cos \theta} = \frac{16 + 9(1 - \cos^2 \theta)}{5 - 3\cos \theta}\) | M1 |
| (b) | \(= \frac{(5 - 3\cos \theta)(5 + 3\cos \theta)}{5 - 3\cos \theta}\) | A1 |
| (b) | \(= 5 + 3\cos\theta\) | A1 |
| (b) | Least possible value is 2 and occurs at \(\theta = \pi\) | B1F |
| Total | 4 |
(a)(i) | $\frac{4\sin x}{\cos x} + \frac{5\cos x}{\cos x} = 0; \quad 4\tan x + 5 = 0$ | M1 | $\frac{\sin x}{\cos x} = \tan x$ clearly used to obtain a linear equation in $\tan x$. $-1.25$ OE. NMS mark as B2 or B0
(a)(i) | $\tan x = -\frac{5}{4}$ $(= -1.25)$ | A1 | 2 marks |
(a)(ii) | $\tan x = 1, \quad \tan x = -1.25$ | B1F | 1 and c's answer to (a)(i) vals for tan (x). PI by a correct angle for both tan values.
(a)(ii) | $(x =) 45°, 225°, 129°, 309°$ | B2, 1 | 3 marks | B2 45, 225, AWRT 129, AWRT 309. If not B2 award B1 for at least two correct. If more than four values in given interval, deduct 1 mark for each extra to min of B0. Ignore values outside $0° \leq x \leq 360°$
| **Total** | **9** |
(b) | $\frac{16 + 9\sin^2 \theta}{5 - 3\cos \theta} = \frac{16 + 9(1 - \cos^2 \theta)}{5 - 3\cos \theta}$ | M1 | Replacing $\sin^2 \theta$ by $1 - \cos^2 \theta$ in given expression or replacing $\cos^2 \theta$ by $1 - \sin^2 \theta$ in term $\pm 3q\cos^2 \theta$.
(b) | $= \frac{(5 - 3\cos \theta)(5 + 3\cos \theta)}{5 - 3\cos \theta}$ | A1 | Or any two of: $5p - 3q = 16, \quad 5q - 3p = 0, \quad 3q = 9$
(b) | $= 5 + 3\cos\theta$ | A1 | CSO. Or q=3, p=5 and checking remaining eqn is satisfied.
(b) | Least possible value is 2 and occurs at $\theta = \pi$ | B1F | Ft on c's p and q non zero values. {If q>0, least val=p−q $\theta = \pi$} {If q<0, least val=p+q at $\theta = 0$} Ignore values of $\theta$ outside given interval
| **Total** | **4** |
**Note on (a)(i):** $\tan x = -1.25$ OE with no errors seen scores 2 marks. Methods involving squaring and $\tan x \neq -1.25$ OE, must give reasons for discounting certain solns before M1A0 scored
**Note on (b):** Multiplying the numerator and denominator of the given expression by $5 + 3\cos\theta$ does not score until the correct relevant identity has been used (M1); the 1st A1 will then be awarded when the rational expression has been written in a correct form with terms which can be cancelled legitimately e.g. $\frac{(16 + 9\sin^2\theta)(5 + 3\cos\theta)}{(16 + 9\sin^2\theta)}$
---8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $4 \sin x + 5 \cos x = 0$, find the value of $\tan x$.
\item Hence solve the equation $( 1 - \tan x ) ( 4 \sin x + 5 \cos x ) = 0$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$, giving your values of $x$ to the nearest degree.
\end{enumerate}\item By first showing that $\frac { 16 + 9 \sin ^ { 2 } \theta } { 5 - 3 \cos \theta }$ can be expressed in the form $p + q \cos \theta$, where $p$ and $q$ are integers, find the least possible value of $\frac { 16 + 9 \sin ^ { 2 } \theta } { 5 - 3 \cos \theta }$.
State the exact value of $\theta$, in radians in the interval $0 \leqslant \theta < 2 \pi$, at which this least value occurs.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2016 Q8 [9]}}