| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.3 Part (a) is a standard algebraic identity proof using difference of squares and the fundamental identity sec²θ - tan²θ = 1. Part (b) applies the result with a substitution (θ = 2α) and solves a straightforward equation. This requires competent manipulation of reciprocal trig functions but follows predictable patterns with no novel insight needed—slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Factorise or obtain an expression in \(\cos^2\theta\) or \(\sec^2\theta\) | M1 | \((\sec^2\theta - \tan^2\theta)(\sec^2\theta + \tan^2\theta)\) or \(-1+\frac{2}{\cos^2\theta}\), OE |
| Use of \(1+\tan^2\theta = \sec^2\theta\) (anywhere) | M1 | The 2nd M1 is for expanding \((1+\tan^2\theta)^2 - \tan^4\theta\) |
| Obtain \(1\times(1+\tan^2\theta+\tan^2\theta) = 1+2\tan^2\theta\) | A1 | Obtain given answer from full and correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Form an equation in \(\tan 2\alpha\), or multiply through by \(\cos^4 2\alpha\) to form an equation in \(\sin 2\alpha\) or \(\cos 2\alpha\) | M1 | \(1+2\tan^2 2\alpha = 2\tan^2 2\alpha(1+\tan^2 2\alpha)\); \(\cos^4 2\alpha + 2\sin^2 2\alpha\cos^2 2\alpha = 2\sin^2 2\alpha\); \(\Rightarrow \sin^4 2\alpha + 2\sin^2 2\alpha - 1 = 0\) or \(\cos^4 2\alpha - 4\cos^2 2\alpha + 2 = 0\) |
| Solve for \(\tan 2\alpha\) or equivalent | M1 | \(\tan 2\alpha = \pm\sqrt{\frac{1}{\sqrt{2}}}\) |
| Obtain one correct solution for \(\alpha\), e.g. \(20.0(30...)°\) | A1 | |
| Obtain a second correct value for \(\alpha\), e.g. \(70.0°\,(69.9698...°)\) | A1 | |
| Obtain solutions \(110°\,(110.0)\) and \(160°\,(160.0)\) for \(\alpha\), and no others in range | A1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Factorise or obtain an expression in $\cos^2\theta$ or $\sec^2\theta$ | M1 | $(\sec^2\theta - \tan^2\theta)(\sec^2\theta + \tan^2\theta)$ or $-1+\frac{2}{\cos^2\theta}$, OE |
| Use of $1+\tan^2\theta = \sec^2\theta$ (anywhere) | M1 | The 2nd M1 is for expanding $(1+\tan^2\theta)^2 - \tan^4\theta$ |
| Obtain $1\times(1+\tan^2\theta+\tan^2\theta) = 1+2\tan^2\theta$ | A1 | Obtain given answer from full and correct working |
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## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Form an equation in $\tan 2\alpha$, or multiply through by $\cos^4 2\alpha$ to form an equation in $\sin 2\alpha$ or $\cos 2\alpha$ | M1 | $1+2\tan^2 2\alpha = 2\tan^2 2\alpha(1+\tan^2 2\alpha)$; $\cos^4 2\alpha + 2\sin^2 2\alpha\cos^2 2\alpha = 2\sin^2 2\alpha$; $\Rightarrow \sin^4 2\alpha + 2\sin^2 2\alpha - 1 = 0$ or $\cos^4 2\alpha - 4\cos^2 2\alpha + 2 = 0$ |
| Solve for $\tan 2\alpha$ or equivalent | M1 | $\tan 2\alpha = \pm\sqrt{\frac{1}{\sqrt{2}}}$ |
| Obtain one correct solution for $\alpha$, e.g. $20.0(30...)°$ | A1 | |
| Obtain a second correct value for $\alpha$, e.g. $70.0°\,(69.9698...°)$ | A1 | |
| Obtain solutions $110°\,(110.0)$ and $160°\,(160.0)$ for $\alpha$, and no others in range | A1 | |
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\begin{enumerate}[label=(\alph*)]
\item Show that $\sec ^ { 4 } \theta - \tan ^ { 4 } \theta \equiv 1 + 2 \tan ^ { 2 } \theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-07_2723_35_101_20}
\item Hence or otherwise solve the equation $\sec ^ { 4 } 2 \alpha - \tan ^ { 4 } 2 \alpha = 2 \tan ^ { 2 } 2 \alpha \sec ^ { 2 } 2 \alpha$ for $0 ^ { \circ } < \alpha < 180 ^ { \circ }$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q4 [8]}}