| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.3 Part (a) requires standard algebraic manipulation using difference of squares and double angle formulas (cos 2θ = cos²θ - sin²θ, sin 2θ = 2sinθcosθ), which are routine A-level techniques. Part (b) applies the proven identity to solve a trigonometric equation in a restricted domain, requiring substitution and solving a quadratic in cos 2α. Both parts are straightforward applications of standard methods with no novel insight required, making this slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rewrite \(\cos^4\theta\) as \(\left(\dfrac{1+\cos 2\theta}{2}\right)^2\) or \(\sin^4\theta\) as \(\left(\dfrac{1-\cos 2\theta}{2}\right)^2\) or \(4\sin^2\theta\cos^2\theta\) as \(\sin^2 2\theta\) | B1 | Starting on left; double angle for one term |
| Obtain \(\left(\dfrac{1+\cos 2\theta}{2}\right)^2 - \left(\dfrac{1-\cos 2\theta}{2}\right)^2 - \sin^2 2\theta\) | B1 | OE, e.g. \(1\times\cos 2\theta - \sin^2 2\theta\) |
| Expand and simplify to obtain \(\cos^2 2\theta + \cos 2\theta - 1\) | B1 | AG |
| Alternative Method 1: Express \(\cos^4\theta - \sin^4\theta\) as \((\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)\) or rewrite \(4\sin^2\theta\cos^2\theta\) as \(\sin^2 2\theta\) | B1 | Starting on left |
| Simplify to \(\cos 2\theta - \sin^2 2\theta\) | B1 | If \(\cos^4\theta - \sin^4\theta = \cos 2\theta\) instead of \((\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta) = \cos 2\theta\), B0 |
| Use \(\sin^2 2\theta = 1 - \cos^2 2\theta\) to obtain \(\cos^2 2\theta + \cos 2\theta - 1\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct double angle formulae once e.g. replace \(\cos 2\theta\) with \(\cos^2\theta - \sin^2\theta\): \((\cos^2\theta - \sin^2\theta)^2 + (\cos^2\theta - \sin^2\theta) - 1\) | B1 | Starting on right. Double angle for one term. |
| Expand to obtain \(\cos^4\theta - 2\sin^2\theta\cos^2\theta - \sin^4\theta + 2\sin^4\theta + \cos^2\theta - \sin^2\theta - 1\) | B1 | Write \(\sin^4\theta\) as \(-\sin^4\theta + 2\sin^4\theta\). Write \(2\sin^2\theta\) as \(2\sin^2\theta(\cos^2\theta + \sin^2\theta)\). |
| Rewrite as \(\cos^4\theta - 2\sin^2\theta\cos^2\theta - \sin^4\theta + 2\sin^4\theta - 2\sin^2\theta\) leading to \(\cos^4\theta - 4\sin^2\theta\cos^2\theta - \sin^4\theta\) | B1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State a quadratic equation in \(\cos 2\alpha\) and solve for \(\alpha\): \((\cos^2 2\alpha + \cos 2\alpha - 1 = 0)\) | M1 | Alternative: form a quadratic in \(\tan^2\alpha\) and solve for \(\alpha\) \((\tan^4\alpha + 4\tan^2\alpha - 1 = 0)\) |
| Obtain \(\alpha = 25.9°\) or \(\alpha = 154.1°\) | A1 | May be more accurate. Allow 154 for 154.1. |
| Obtain \(\alpha = 25.9°\) and \(\alpha = 154.1°\) and no others in range | A1 | May be more accurate. Allow 154 for 154.1. Mark answers in radians as a misread \((0.452, 2.69)\). |
| Total | 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rewrite $\cos^4\theta$ as $\left(\dfrac{1+\cos 2\theta}{2}\right)^2$ or $\sin^4\theta$ as $\left(\dfrac{1-\cos 2\theta}{2}\right)^2$ or $4\sin^2\theta\cos^2\theta$ as $\sin^2 2\theta$ | B1 | Starting on left; double angle for one term |
| Obtain $\left(\dfrac{1+\cos 2\theta}{2}\right)^2 - \left(\dfrac{1-\cos 2\theta}{2}\right)^2 - \sin^2 2\theta$ | B1 | OE, e.g. $1\times\cos 2\theta - \sin^2 2\theta$ |
| Expand and simplify to obtain $\cos^2 2\theta + \cos 2\theta - 1$ | B1 | AG |
| **Alternative Method 1:** Express $\cos^4\theta - \sin^4\theta$ as $(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)$ or rewrite $4\sin^2\theta\cos^2\theta$ as $\sin^2 2\theta$ | B1 | Starting on left |
| Simplify to $\cos 2\theta - \sin^2 2\theta$ | B1 | If $\cos^4\theta - \sin^4\theta = \cos 2\theta$ instead of $(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta) = \cos 2\theta$, B0 |
| Use $\sin^2 2\theta = 1 - \cos^2 2\theta$ to obtain $\cos^2 2\theta + \cos 2\theta - 1$ | B1 | AG |
## Question 5(a) — Alternative Method 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct double angle formulae once e.g. replace $\cos 2\theta$ with $\cos^2\theta - \sin^2\theta$: $(\cos^2\theta - \sin^2\theta)^2 + (\cos^2\theta - \sin^2\theta) - 1$ | B1 | Starting on right. Double angle for one term. |
| Expand to obtain $\cos^4\theta - 2\sin^2\theta\cos^2\theta - \sin^4\theta + 2\sin^4\theta + \cos^2\theta - \sin^2\theta - 1$ | B1 | Write $\sin^4\theta$ as $-\sin^4\theta + 2\sin^4\theta$. Write $2\sin^2\theta$ as $2\sin^2\theta(\cos^2\theta + \sin^2\theta)$. |
| Rewrite as $\cos^4\theta - 2\sin^2\theta\cos^2\theta - \sin^4\theta + 2\sin^4\theta - 2\sin^2\theta$ leading to $\cos^4\theta - 4\sin^2\theta\cos^2\theta - \sin^4\theta$ | B1 | |
| **Total** | **3** | |
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## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State a quadratic equation in $\cos 2\alpha$ and solve for $\alpha$: $(\cos^2 2\alpha + \cos 2\alpha - 1 = 0)$ | M1 | Alternative: form a quadratic in $\tan^2\alpha$ and solve for $\alpha$ $(\tan^4\alpha + 4\tan^2\alpha - 1 = 0)$ |
| Obtain $\alpha = 25.9°$ or $\alpha = 154.1°$ | A1 | May be more accurate. Allow 154 for 154.1. |
| Obtain $\alpha = 25.9°$ and $\alpha = 154.1°$ and no others in range | A1 | May be more accurate. Allow 154 for 154.1. Mark answers in radians as a misread $(0.452, 2.69)$. |
| **Total** | **3** | |
---5
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos ^ { 4 } \theta - \sin ^ { 4 } \theta - 4 \sin ^ { 2 } \theta \cos ^ { 2 } \theta \equiv \cos ^ { 2 } 2 \theta + \cos 2 \theta - 1$.
\item Solve the equation $\cos ^ { 4 } \alpha - \sin ^ { 4 } \alpha = 4 \sin ^ { 2 } \alpha \cos ^ { 2 } \alpha$ for $0 ^ { \circ } \leqslant \alpha \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q5 [6]}}