| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Integration using reciprocal identities |
| Difficulty | Standard +0.8 This question requires proving a non-standard trigonometric identity using double-angle formulas and reciprocal definitions, then applying it to evaluate a definite integral. Part (a) demands algebraic manipulation beyond routine recall, while part (b) requires recognizing that tan θ integrates to -ln|cos θ| and careful evaluation at the given limits. The combination of proof and integration with specific bounds elevates this above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express the LHS in terms of \(\cos 2\theta\) and \(\sin 2\theta\) | B1 | e.g. \(\frac{1}{\sin 2\theta} - \frac{\cos 2\theta}{\sin 2\theta}\) |
| Use correct double angle formulae to express the LHS in terms of \(\cos\theta\) and \(\sin\theta\) | M1 | e.g. \(\frac{1-(1-2\sin^2\theta)}{2\sin\theta\cos\theta}\) |
| Obtain \(\tan\theta\) from correct working | A1 | AG |
| Alternative method 1: Express LHS in terms of \(\sin 2\theta\) and \(\tan 2\theta\) | B1 | |
| Use correct double angle formulae to express LHS in terms of \(\cos\theta\) and \(\sin\theta\) | M1 | e.g. \(\frac{1}{2\sin\theta\cos\theta} - \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{\frac{2\sin\theta}{\cos\theta}}\left(=\frac{4\sin^2\theta}{4\sin\theta\cos\theta}\right)\) |
| Obtain \(\tan\theta\) from correct working | A1 | AG |
| Alternative method 2: Express LHS in terms of \(\sin 2\theta\) and \(\tan 2\theta\) | B1 | |
| Use correct \(t\) substitution or rearrangement of \(\sin 2\theta\) in terms of \(\sec^2 2\theta\) and \(\tan\theta\) to express LHS in terms of \(\tan\theta\) | M1 | \(\left(\frac{\sec^2\theta}{2\tan\theta} - \frac{1-\tan^2\theta}{2\tan\theta} = \right)\frac{1+\tan^2}{2\tan} - \frac{1-\tan^2}{2\tan}\) |
| Obtain \(\tan\theta\) from correct working | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State integral of the form \(\mp\ln\cos\theta\) or \(\pm\ln\sec\theta\) | \*M1 | \([-\ln\cos\theta]^{\frac{\pi}{4}}_{\frac{\pi}{8}}\) OE |
| Use correct limits correctly and insert exact values for the trigonometric ratios | DM1 | Need to see evidence of the substitution |
| Obtain a correct expression, e.g. \(-\ln\frac{1}{2}+\ln\frac{1}{\sqrt{2}}\) | A1 | |
| Obtain \(\frac{1}{2}\ln 2\) from correct working | A1 | AG (must see an intermediate step) |
| Total | 4 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Express the LHS in terms of $\cos 2\theta$ and $\sin 2\theta$ | B1 | e.g. $\frac{1}{\sin 2\theta} - \frac{\cos 2\theta}{\sin 2\theta}$ |
| Use correct double angle formulae to express the LHS in terms of $\cos\theta$ and $\sin\theta$ | M1 | e.g. $\frac{1-(1-2\sin^2\theta)}{2\sin\theta\cos\theta}$ |
| Obtain $\tan\theta$ from correct working | A1 | AG |
| **Alternative method 1:** Express LHS in terms of $\sin 2\theta$ and $\tan 2\theta$ | B1 | |
| Use correct double angle formulae to express LHS in terms of $\cos\theta$ and $\sin\theta$ | M1 | e.g. $\frac{1}{2\sin\theta\cos\theta} - \frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{\frac{2\sin\theta}{\cos\theta}}\left(=\frac{4\sin^2\theta}{4\sin\theta\cos\theta}\right)$ |
| Obtain $\tan\theta$ from correct working | A1 | AG |
| **Alternative method 2:** Express LHS in terms of $\sin 2\theta$ and $\tan 2\theta$ | B1 | |
| Use correct $t$ substitution or rearrangement of $\sin 2\theta$ in terms of $\sec^2 2\theta$ and $\tan\theta$ to express LHS in terms of $\tan\theta$ | M1 | $\left(\frac{\sec^2\theta}{2\tan\theta} - \frac{1-\tan^2\theta}{2\tan\theta} = \right)\frac{1+\tan^2}{2\tan} - \frac{1-\tan^2}{2\tan}$ |
| Obtain $\tan\theta$ from correct working | A1 | AG |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State integral of the form $\mp\ln\cos\theta$ or $\pm\ln\sec\theta$ | \*M1 | $[-\ln\cos\theta]^{\frac{\pi}{4}}_{\frac{\pi}{8}}$ OE |
| Use correct limits correctly and insert exact values for the trigonometric ratios | DM1 | Need to see evidence of the substitution |
| Obtain a correct expression, e.g. $-\ln\frac{1}{2}+\ln\frac{1}{\sqrt{2}}$ | A1 | |
| Obtain $\frac{1}{2}\ln 2$ from correct working | A1 | AG (must see an intermediate step) |
| **Total** | **4** | |
---6
\begin{enumerate}[label=(\alph*)]
\item Prove that $\operatorname { cosec } 2 \theta - \cot 2 \theta \equiv \tan \theta$.
\item Hence show that $\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 3 } \pi } ( \operatorname { cosec } 2 \theta - \cot 2 \theta ) \mathrm { d } \theta = \frac { 1 } { 2 } \ln 2$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q6 [7]}}