| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Integration using reciprocal identities |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard double-angle identities (cos 2θ = 1 - 2sin²θ = 2cos²θ - 1) to prove the identity, then integrating tan²θ using the identity tan²θ = sec²θ - 1. Both techniques are routine for P3/C3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct double angle formula or \(t\)-substitution twice | M1 | |
| Obtain \(\dfrac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2\theta\) from correct working | A1 | AG |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Express \(\tan^2\theta\) in terms of \(\sec^2\theta\) | M1 | \(\left(\displaystyle\int_{\pi/6}^{\pi/3}(\sec^2\theta \pm 1)\,d\theta\right)\) |
| Integrate and obtain terms \(\tan\theta - \theta\) | A1 | Accept with a mixture of \(x\) and \(\theta\) |
| Substitute limits correctly in an integral of the form \(a\tan\theta + b\theta\), where \(ab \neq 0\) | M1 | \(\left(\sqrt{3} - \dfrac{\pi}{3} - \dfrac{1}{\sqrt{3}} + \dfrac{\pi}{6}\right)\) Allow if trig. not substituted |
| Obtain answer \(\dfrac{2}{3}\sqrt{3} - \dfrac{1}{6}\pi\) | A1 | or equivalent exact 2-term expression |
| Total | 4 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct double angle formula or $t$-substitution twice | M1 | |
| Obtain $\dfrac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2\theta$ from correct working | A1 | AG |
| **Total** | **2** | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Express $\tan^2\theta$ in terms of $\sec^2\theta$ | M1 | $\left(\displaystyle\int_{\pi/6}^{\pi/3}(\sec^2\theta \pm 1)\,d\theta\right)$ |
| Integrate and obtain terms $\tan\theta - \theta$ | A1 | Accept with a mixture of $x$ and $\theta$ |
| Substitute limits correctly in an integral of the form $a\tan\theta + b\theta$, where $ab \neq 0$ | M1 | $\left(\sqrt{3} - \dfrac{\pi}{3} - \dfrac{1}{\sqrt{3}} + \dfrac{\pi}{6}\right)$ Allow if trig. not substituted |
| Obtain answer $\dfrac{2}{3}\sqrt{3} - \dfrac{1}{6}\pi$ | A1 | or equivalent exact 2-term expression |
| **Total** | **4** | |
---4
\begin{enumerate}[label=(\alph*)]
\item Prove that $\frac { 1 - \cos 2 \theta } { 1 + \cos 2 \theta } \equiv \tan ^ { 2 } \theta$.
\item Hence find the exact value of $\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \frac { 1 - \cos 2 \theta } { 1 + \cos 2 \theta } \mathrm {~d} \theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q4 [6]}}