| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Challenging +1.2 This is a multi-part question requiring double angle identities, reciprocal trig functions, equation solving, and integration. Part (i) is a standard 'show that' identity proof. Part (ii) requires substituting the identity and solving a quadratic in tan α, which is moderately challenging. Part (iii) involves recognizing a substitution or algebraic manipulation for integration. While requiring multiple techniques across three parts, each component uses standard A-level methods without requiring exceptional insight, placing it moderately above average difficulty. |
| 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^21.05l Double angle formulae: and compound angle formulae1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\cos 2\theta = 2\cos^2\theta - 1\) appropriately twice | B1 | Alternative method: \(\dfrac{1-2\sin^2\theta}{2\cos^2\theta} = \dfrac{1}{2}\sec^2\theta - \tan^2\theta\) or \(\dfrac{1}{2\cos^2\theta} - \tan^2\theta\) — B1 |
| Simplify to confirm \(1 - \frac{1}{2}\sec^2\theta\) | B1 | then as for 2nd B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\sec^2\alpha = 1 + \tan^2\alpha\) | B1 | |
| Obtain equation \(\tan^2\alpha + 10\tan\alpha + 25 = 0\) or equivalent | B1 | |
| Attempt solution of 3-term quadratic for \(\tan\alpha\) and use correct process for finding value of \(\alpha\) from negative value of \(\tan\alpha\) | M1 | If quadratic is incorrect, need to see evidence of attempt to solve as required to obtain M1 |
| Obtain \(1.77\) | A1 | Allow better or in terms of \(\pi\) \(\left(\dfrac{1013\pi}{1800}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply integrand \(1 - \frac{1}{2}\sec^2\frac{1}{2}x\) | B1 | |
| Obtain integral of form \(k_1 x - k_2\tan\frac{1}{2}x\) | M1 | |
| Obtain correct \(x - \tan\frac{1}{2}x\) | A1 | |
| Apply limits correctly to obtain \(\pi - 2\) | A1 |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\cos 2\theta = 2\cos^2\theta - 1$ appropriately twice | B1 | Alternative method: $\dfrac{1-2\sin^2\theta}{2\cos^2\theta} = \dfrac{1}{2}\sec^2\theta - \tan^2\theta$ or $\dfrac{1}{2\cos^2\theta} - \tan^2\theta$ — B1 |
| Simplify to confirm $1 - \frac{1}{2}\sec^2\theta$ | B1 | then as for 2nd B1 |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\sec^2\alpha = 1 + \tan^2\alpha$ | B1 | |
| Obtain equation $\tan^2\alpha + 10\tan\alpha + 25 = 0$ or equivalent | B1 | |
| Attempt solution of 3-term quadratic for $\tan\alpha$ and use correct process for finding value of $\alpha$ from negative value of $\tan\alpha$ | M1 | If quadratic is incorrect, need to see evidence of attempt to solve as required to obtain M1 |
| Obtain $1.77$ | A1 | Allow better or in terms of $\pi$ $\left(\dfrac{1013\pi}{1800}\right)$ |
## Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply integrand $1 - \frac{1}{2}\sec^2\frac{1}{2}x$ | B1 | |
| Obtain integral of form $k_1 x - k_2\tan\frac{1}{2}x$ | M1 | |
| Obtain correct $x - \tan\frac{1}{2}x$ | A1 | |
| Apply limits correctly to obtain $\pi - 2$ | A1 | |6 (i) Show that $\frac { \cos 2 \theta } { 1 + \cos 2 \theta } \equiv 1 - \frac { 1 } { 2 } \sec ^ { 2 } \theta$.\\
(ii) Solve the equation $\frac { \cos 2 \alpha } { 1 + \cos 2 \alpha } = 13 + 5 \tan \alpha$ for $0 < \alpha < \pi$.\\
(iii) Find the exact value of $\int _ { - \frac { 1 } { 2 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \cos x } { 1 + \cos x } \mathrm {~d} x$.
\hfill \mbox{\textit{CAIE P2 2016 Q6 [10]}}