| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) is a standard algebraic manipulation of trig identities requiring common denominator work and Pythagorean identity. Part (b) uses the result to solve an equation, leading to a quadratic in cos θ with exact values. This is routine A-level Further Maths content with straightforward steps, slightly easier than average due to the 'hence' structure guiding the solution. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{LHS}=\frac{\sin^2\theta-(1-\cos\theta)}{(1-\cos\theta)\sin\theta}\) | M1 | Attempt to write LHS as a single fraction; where candidates manipulate entire statement, allow M1 for eliminating or combining fractions e.g. multiplying through by \((1-\cos\theta)\sin\theta\) |
| \(=\frac{(1-\cos^2\theta)+\cos\theta-1}{(1-\cos\theta)\sin\theta}\) | B1 | Use of identity \(\sin^2\theta=1-\cos^2\theta\); M1 for algebraic manipulation leading to a known identity, B1 identity obtained, A1 complete proof |
| \(=\frac{\cos\theta(1-\cos\theta)}{\sin\theta(1-\cos\theta)}\) | M1 | Algebraic manipulation e.g. factorising the numerator |
| \(=\cot\theta\) | A1 [4] | AG Complete proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin\theta}{(1-\cos\theta)}\cdot\frac{(1+\cos\theta)}{(1+\cos\theta)}=\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}\) | M1 | Attempt to change the denominator of the fraction |
| B1 | Use of trig identity | |
| \(\frac{(1+\cos\theta)}{\sin\theta}-\frac{1}{\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta\) | M1 | Combining the fractions |
| A1 | AG Complete proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses \(\frac{1}{\tan\theta}=3\tan\theta\) | M1 | soi; also allow for equivalent equation in \(\cot\theta\) |
| \(\tan\theta=\pm\frac{1}{\sqrt{3}}\) | M1 | Method must be clearly using the given answer in (a); allow positive root only for second M mark |
| \(\theta=\frac{\pi}{6},\frac{5\pi}{6}\) | A1 [3] | Do not allow if additional answers in the interval; ignore additional values outside the interval |
## Question 6:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{LHS}=\frac{\sin^2\theta-(1-\cos\theta)}{(1-\cos\theta)\sin\theta}$ | M1 | Attempt to write LHS as a single fraction; where candidates manipulate entire statement, allow M1 for eliminating or combining fractions e.g. multiplying through by $(1-\cos\theta)\sin\theta$ |
| $=\frac{(1-\cos^2\theta)+\cos\theta-1}{(1-\cos\theta)\sin\theta}$ | B1 | Use of identity $\sin^2\theta=1-\cos^2\theta$; M1 for algebraic manipulation leading to a known identity, B1 identity obtained, A1 complete proof |
| $=\frac{\cos\theta(1-\cos\theta)}{\sin\theta(1-\cos\theta)}$ | M1 | Algebraic manipulation e.g. factorising the numerator |
| $=\cot\theta$ | A1 [4] | AG Complete proof |
**Alternative solution:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin\theta}{(1-\cos\theta)}\cdot\frac{(1+\cos\theta)}{(1+\cos\theta)}=\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}$ | M1 | Attempt to change the denominator of the fraction |
| | B1 | Use of trig identity |
| $\frac{(1+\cos\theta)}{\sin\theta}-\frac{1}{\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta$ | M1 | Combining the fractions |
| | A1 | AG Complete proof |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses $\frac{1}{\tan\theta}=3\tan\theta$ | M1 | soi; also allow for equivalent equation in $\cot\theta$ |
| $\tan\theta=\pm\frac{1}{\sqrt{3}}$ | M1 | Method must be clearly using the given answer in (a); allow positive root only for second M mark |
| $\theta=\frac{\pi}{6},\frac{5\pi}{6}$ | A1 [3] | Do not allow if additional answers in the interval; ignore additional values outside the interval |
---6
\begin{enumerate}[label=(\alph*)]
\item Prove that $\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } = \cot \theta$.
\item Hence find the exact roots of the equation $\frac { \sin \theta } { 1 - \cos \theta } - \frac { 1 } { \sin \theta } = 3 \tan \theta$ in the interval $0 \leqslant \theta \leqslant \pi$.
Answer all the questions.\\
Section B (75 marks)
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q6 [7]}}