| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation of trig identities (converting to sin/cos, simplifying) followed by a routine equation solve. The identity proof follows a predictable path using basic identities, and part (ii) is a direct substitution leading to a standard cos equation. Slightly above average due to the algebraic manipulation required, but well within typical A-level scope with no novel insights needed. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)^2 = \left(\frac{1}{s} - \frac{c}{s}\right)^2\) | M1 | Use of \(\tan = \sin/\cos\) |
| \(\frac{(1-c)^2}{s^2} = \frac{(1-c)^2}{1-c^2}\) | M1 | Use of \(s^2 = 1-c^2\) |
| \(= \frac{(1-c)(1-c)}{(1-c)(1+c)}\) or \(\frac{(1-c)^2}{(1-c)(1+c)}\) | A1 | |
| \(\equiv \frac{1-\cos x}{1+\cos x}\) | A1 [4] | ag |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)^2 = \frac{2}{5}\) | ||
| \(\frac{1-\cos x}{1+\cos x} = \frac{2}{5} \rightarrow \cos x = \frac{3}{7}\) | M1 | Making \(\cos x\) the subject |
| \(\rightarrow x = 1.13\) or \(5.16\) | A1 A1\(^\checkmark\) [3] | \(2\pi - 1^\text{st}\) answer |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)^2 = \left(\frac{1}{s} - \frac{c}{s}\right)^2$ | **M1** | Use of $\tan = \sin/\cos$ |
| $\frac{(1-c)^2}{s^2} = \frac{(1-c)^2}{1-c^2}$ | **M1** | Use of $s^2 = 1-c^2$ |
| $= \frac{(1-c)(1-c)}{(1-c)(1+c)}$ or $\frac{(1-c)^2}{(1-c)(1+c)}$ | **A1** | |
| $\equiv \frac{1-\cos x}{1+\cos x}$ | **A1** [4] | ag |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)^2 = \frac{2}{5}$ | | |
| $\frac{1-\cos x}{1+\cos x} = \frac{2}{5} \rightarrow \cos x = \frac{3}{7}$ | **M1** | Making $\cos x$ the subject |
| $\rightarrow x = 1.13$ or $5.16$ | **A1 A1$^\checkmark$** [3] | $2\pi - 1^\text{st}$ answer |
---4 (i) Prove the identity $\left( \frac { 1 } { \sin x } - \frac { 1 } { \tan x } \right) ^ { 2 } \equiv \frac { 1 - \cos x } { 1 + \cos x }$.\\
(ii) Hence solve the equation $\left( \frac { 1 } { \sin x } - \frac { 1 } { \tan x } \right) ^ { 2 } = \frac { 2 } { 5 }$ for $0 \leqslant x \leqslant 2 \pi$.
\hfill \mbox{\textit{CAIE P1 2015 Q4 [7]}}