| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: algebraic fraction manipulation |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric equation requiring algebraic manipulation (combining fractions, using sin²θ + cos²θ = 1) followed by solving a quadratic in sin θ. The techniques are routine for P1 level with no novel insight required, making it slightly easier than average but not trivial due to the multi-step algebraic manipulation. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin\theta-\cos\theta+\sin\theta+\cos\theta}{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}\left[=\frac{\sin\theta-\cos\theta+\sin\theta+\cos\theta}{\sin^2\theta-\cos^2\theta}\right]=1\) | \*M1 | Use common denominator and equate to 1. |
| \(2\sin\theta[=\sin^2\theta-\cos^2\theta]=\sin^2\theta-(1-\sin^2\theta)\) | DM1 | Multiply by common denominator and replace \(\cos^2\theta\) by \(1-\sin^2\theta\). |
| \(2\sin^2\theta-2\sin\theta-1=0\) | A1 | OE In the given form. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\sin\theta=]\frac{2\pm\sqrt{(-2)^2-4(2)(-1)}}{4}\left[=\frac{2\pm\sqrt{4+8}}{4}=\frac{1\pm\sqrt{3}}{2}\right]\) | M1 | Use formula or complete the square to solve a quadratic equation of the correct form. |
| \(201.5°\) or \(338.5°\) | A1 A1 FT | AWRT; A1 for either solution correct. A1 FT for \(540-(\) first value\()\). If M0, allow SC B1 B1FT similarly. |
| 3 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin\theta-\cos\theta+\sin\theta+\cos\theta}{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}\left[=\frac{\sin\theta-\cos\theta+\sin\theta+\cos\theta}{\sin^2\theta-\cos^2\theta}\right]=1$ | **\*M1** | Use common denominator and equate to 1. |
| $2\sin\theta[=\sin^2\theta-\cos^2\theta]=\sin^2\theta-(1-\sin^2\theta)$ | **DM1** | Multiply by common denominator and replace $\cos^2\theta$ by $1-\sin^2\theta$. |
| $2\sin^2\theta-2\sin\theta-1=0$ | **A1** | OE In the given form. |
| | **3** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\sin\theta=]\frac{2\pm\sqrt{(-2)^2-4(2)(-1)}}{4}\left[=\frac{2\pm\sqrt{4+8}}{4}=\frac{1\pm\sqrt{3}}{2}\right]$ | **M1** | Use formula or complete the square to solve a quadratic equation of the correct form. |
| $201.5°$ or $338.5°$ | **A1 A1 FT** | AWRT; A1 for either solution correct. A1 FT for $540-($ first value$)$. If M0, allow **SC B1 B1FT** similarly. |
| | **3** | |
---6
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$$
may be expressed in the form $a \sin ^ { 2 } \theta + b \sin \theta + c = 0$, where $a$, $b$ and $c$ are constants to be found.
\item Hence solve the equation $\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q6 [6]}}