| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) requires algebraic manipulation of trigonometric fractions with a common denominator approach and use of sin²θ + cos²θ = 1, which is standard P1 technique. Part (b) is a straightforward application using the proven identity to create a simple equation. The algebra is moderately involved but follows routine procedures without requiring novel insight. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin^3\theta}{\sin\theta-1} - \frac{\sin^2\theta}{1+\sin\theta} = \frac{\sin^3\theta(1+\sin\theta)}{(\sin\theta-1)(1+\sin\theta)} - \frac{\sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}\) | *M1 | Using a common denominator |
| \(-\frac{\sin^2\theta + \sin^4\theta}{1-\sin^2\theta}\) | DM1 | Reaching \(\pm(1-\sin^2\theta)\) in denominator. SOI by \(\pm\cos^2\theta\) |
| \(-\frac{\sin^2\theta(1+\sin^2\theta)}{\cos^2\theta}\) | DM1 | Using \(\sin^2\theta + \cos^2\theta = 1\) in denominator and isolating \(\sin^2\theta\) in numerator |
| \(-\tan^2\theta(1+\sin^2\theta)\) | A1 | AG. Using/stating \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) is sufficient for A1. May be working from both sides provided argument is complete. A0 if \(\theta\) or brackets missing throughout, or sign errors. Allow recovery if AG follows from their working |
| \(\frac{\sin^3\theta}{\sin\theta-1} - \frac{\sin^2\theta}{1+\sin\theta}\) | A1 | AG. A0 if \(\theta\) or brackets missing throughout, or sign errors. Allow recovery if AG follows from their working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\tan^2\theta(1+\sin^2\theta) = -\frac{\sin^2\theta(1+\sin^2\theta)}{1-\sin^2\theta}\) | *M1 | Using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta = 1\) |
| \(\frac{-\sin^2\theta - \sin^4\theta}{(1-\sin\theta)(1+\sin\theta)}\) | DM1 | Factorising denominator |
| \(\frac{\sin^2\theta + \sin^3\theta - \sin^3\theta + \sin^4\theta}{(\sin\theta-1)(1+\sin\theta)} = \frac{\sin^3\theta(1+\sin\theta) - \sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}\) | DM1 | Factorising numerator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\tan^2\theta(1+\sin^2\theta) = \tan^2\theta(1-\sin^2\theta)\) leading to \([2]\tan^2\theta = 0\) | M1 | Obtaining a \((\text{trig function})^2 = 0\) WWW |
| \(\tan\theta = 0\) leading to \([\theta =]\ \pi\) | A1 | Ignore extra solutions outside the interval \((0, 2\pi)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-\frac{\sin^2\theta}{\cos^2\theta}(1+\sin^2\theta) = \frac{\sin^2\theta}{\cos^2\theta}(1-\sin^2\theta)\) leading to \(-\sin^2\theta - \sin^4\theta = \sin^2\theta - \sin^4\theta\) leading to \([2]\sin^2\theta = 0\) | M1 | Obtaining a \((\text{trig function})^2 = 0\) WWW |
| \(\sin\theta = 0\) leading to \([\theta =]\ \pi\) | A1 | Ignore extra solutions outside the interval \((0, 2\pi)\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin^3\theta}{\sin\theta-1} - \frac{\sin^2\theta}{1+\sin\theta} = \frac{\sin^3\theta(1+\sin\theta)}{(\sin\theta-1)(1+\sin\theta)} - \frac{\sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}$ | *M1 | Using a common denominator |
| $-\frac{\sin^2\theta + \sin^4\theta}{1-\sin^2\theta}$ | DM1 | Reaching $\pm(1-\sin^2\theta)$ in denominator. SOI by $\pm\cos^2\theta$ |
| $-\frac{\sin^2\theta(1+\sin^2\theta)}{\cos^2\theta}$ | DM1 | Using $\sin^2\theta + \cos^2\theta = 1$ in denominator and isolating $\sin^2\theta$ in numerator |
| $-\tan^2\theta(1+\sin^2\theta)$ | A1 | AG. Using/stating $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is sufficient for A1. May be working from both sides provided argument is complete. A0 if $\theta$ or brackets missing throughout, or sign errors. Allow recovery if AG follows from their working |
| $\frac{\sin^3\theta}{\sin\theta-1} - \frac{\sin^2\theta}{1+\sin\theta}$ | A1 | AG. A0 if $\theta$ or brackets missing throughout, or sign errors. Allow recovery if AG follows from their working |
**Alternative method for Q4(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\tan^2\theta(1+\sin^2\theta) = -\frac{\sin^2\theta(1+\sin^2\theta)}{1-\sin^2\theta}$ | *M1 | Using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ |
| $\frac{-\sin^2\theta - \sin^4\theta}{(1-\sin\theta)(1+\sin\theta)}$ | DM1 | Factorising denominator |
| $\frac{\sin^2\theta + \sin^3\theta - \sin^3\theta + \sin^4\theta}{(\sin\theta-1)(1+\sin\theta)} = \frac{\sin^3\theta(1+\sin\theta) - \sin^2\theta(\sin\theta-1)}{(\sin\theta-1)(1+\sin\theta)}$ | DM1 | Factorising numerator |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\tan^2\theta(1+\sin^2\theta) = \tan^2\theta(1-\sin^2\theta)$ leading to $[2]\tan^2\theta = 0$ | M1 | Obtaining a $(\text{trig function})^2 = 0$ WWW |
| $\tan\theta = 0$ leading to $[\theta =]\ \pi$ | A1 | Ignore extra solutions outside the interval $(0, 2\pi)$ |
**Alternative method for Q4(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-\frac{\sin^2\theta}{\cos^2\theta}(1+\sin^2\theta) = \frac{\sin^2\theta}{\cos^2\theta}(1-\sin^2\theta)$ leading to $-\sin^2\theta - \sin^4\theta = \sin^2\theta - \sin^4\theta$ leading to $[2]\sin^2\theta = 0$ | M1 | Obtaining a $(\text{trig function})^2 = 0$ WWW |
| $\sin\theta = 0$ leading to $[\theta =]\ \pi$ | A1 | Ignore extra solutions outside the interval $(0, 2\pi)$ |4
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } \equiv - \tan ^ { 2 } \theta \left( 1 + \sin ^ { 2 } \theta \right)$.
\item Hence solve the equation
$$\frac { \sin ^ { 3 } \theta } { \sin \theta - 1 } - \frac { \sin ^ { 2 } \theta } { 1 + \sin \theta } = \tan ^ { 2 } \theta \left( 1 - \sin ^ { 2 } \theta \right)$$
for $0 < \theta < 2 \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q4 [6]}}