| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: tan/sin/cos identity manipulation |
| Difficulty | Moderate -0.3 Part (a) requires converting tan θ sin θ = 1 to (sin²θ)/(cos θ) = 1, then forming sin²θ = cos θ, substituting 1-cos²θ to get a quadratic in cos θ, and solving for θ in the given range—a standard multi-step procedure. Part (b) is routine algebraic manipulation of trig identities. Both parts are typical P1-level exercises requiring familiar techniques without novel insight. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta\sin\theta = 1\) leading to \(\sin^2\theta = \cos\theta\) | M1 | Use of \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and multiplication by \(\cos\theta\) |
| \(1 - \cos^2\theta = \cos\theta\) or \(\cos^2\theta + \cos\theta - 1\ [=0]\) | M1 | Use of trig identity to form a 3-term quadratic |
| \([\cos\theta =]\ \frac{-1\pm\sqrt{5}}{2}\) | M1 | Use of formula or completion of the square must be seen on a 3-term quadratic. Expect \(0.6180\) |
| \(51.8°\) | A1 | Both A marks dependent on the 2nd M1 |
| \(308.2°\) | A1 FT | FT for \((360° - \text{1st soln})\), A0 if extra solutions in range. Radians \(0.905\) and \(5.38\), A1 only for both |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\tan\theta}{\sin\theta} - \frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\sin\theta\cos\theta} - \frac{\sin\theta\cos\theta}{\sin\theta} = \frac{1}{\cos\theta} - \cos\theta\) | M1 | Use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) twice with correct use of fractions |
| \(= \frac{1-\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}\) | M1 | Use \(1 - \cos^2\theta = \sin^2\theta\) with correct use of fractions |
| \(= \tan\theta\sin\theta\) | A1 | WWW |
| 3 |
## Question 7(a):
| $\tan\theta\sin\theta = 1$ leading to $\sin^2\theta = \cos\theta$ | M1 | Use of $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and multiplication by $\cos\theta$ |
|---|---|---|
| $1 - \cos^2\theta = \cos\theta$ or $\cos^2\theta + \cos\theta - 1\ [=0]$ | M1 | Use of trig identity to form a 3-term quadratic |
| $[\cos\theta =]\ \frac{-1\pm\sqrt{5}}{2}$ | M1 | Use of formula or completion of the square must be seen on a 3-term quadratic. Expect $0.6180$ |
| $51.8°$ | A1 | Both A marks dependent on the 2nd M1 |
| $308.2°$ | A1 FT | FT for $(360° - \text{1st soln})$, A0 if extra solutions in range. Radians $0.905$ and $5.38$, A1 only for both |
| | **5** | |
## Question 7(b):
| $\frac{\tan\theta}{\sin\theta} - \frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\sin\theta\cos\theta} - \frac{\sin\theta\cos\theta}{\sin\theta} = \frac{1}{\cos\theta} - \cos\theta$ | M1 | Use $\tan\theta = \frac{\sin\theta}{\cos\theta}$ twice with correct use of fractions |
|---|---|---|
| $= \frac{1-\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$ | M1 | Use $1 - \cos^2\theta = \sin^2\theta$ with correct use of fractions |
| $= \tan\theta\sin\theta$ | A1 | WWW |
| | **3** | |
---7
\begin{enumerate}[label=(\alph*)]
\item By first obtaining a quadratic equation in $\cos \theta$, solve the equation
$$\tan \theta \sin \theta = 1$$
for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.
\item Show that $\frac { \tan \theta } { \sin \theta } - \frac { \sin \theta } { \tan \theta } \equiv \tan \theta \sin \theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q7 [8]}}