| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| 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 (i) requires routine algebraic manipulation (common denominator, difference of squares) and standard trig identities (sin²θ + cos²θ = 1, tan θ = sin θ/cos θ). Part (ii) is a straightforward substitution using the proven identity, leading to a simple equation 4cos θ = 3. This is a standard multi-part question with predictable techniques and no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{LHS} = \frac{1+2c+c^2-(1-2c+c^2)}{(1-c)(1+c)}\) | M1 | Attempt at combining fractions |
| \(= \frac{4c}{1-c^2}\) | A1 A1 | A1 for numerator, A1 for denominator |
| \(= \frac{4c}{s^2}\) | Essential step for award of A1 | |
| \(= \frac{4}{ts}\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\theta\!\left(\frac{1+\cos\theta}{1-\cos\theta} - \frac{1-\cos\theta}{1+\cos\theta}\right) = 3\) → \(s \times \frac{4}{ts} = 3\) \(\left(\to t = \frac{4}{3}\right)\) | M1 | Uses part (i) to eliminate "\(s\)" correctly |
| \(\theta = 53.1°\) and \(233.1°\) | A1 A1\(\checkmark\) | \(\checkmark\) for \(180°+\) 1st answer |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{LHS} = \frac{1+2c+c^2-(1-2c+c^2)}{(1-c)(1+c)}$ | M1 | Attempt at combining fractions |
| $= \frac{4c}{1-c^2}$ | A1 A1 | A1 for numerator, A1 for denominator |
| $= \frac{4c}{s^2}$ | | Essential step for award of A1 |
| $= \frac{4}{ts}$ **AG** | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta\!\left(\frac{1+\cos\theta}{1-\cos\theta} - \frac{1-\cos\theta}{1+\cos\theta}\right) = 3$ → $s \times \frac{4}{ts} = 3$ $\left(\to t = \frac{4}{3}\right)$ | M1 | Uses part (i) to eliminate "$s$" correctly |
| $\theta = 53.1°$ and $233.1°$ | A1 A1$\checkmark$ | $\checkmark$ for $180°+$ 1st answer |7 (i) Prove the identity $\frac { 1 + \cos \theta } { 1 - \cos \theta } - \frac { 1 - \cos \theta } { 1 + \cos \theta } \equiv \frac { 4 } { \sin \theta \tan \theta }$.\\
(ii) Hence solve, for $0 ^ { \circ } < \theta < 360 ^ { \circ }$, the equation
$$\sin \theta \left( \frac { 1 + \cos \theta } { 1 - \cos \theta } - \frac { 1 - \cos \theta } { 1 + \cos \theta } \right) = 3 .$$
\hfill \mbox{\textit{CAIE P1 2016 Q7 [7]}}