| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (i) requires algebraic manipulation of trig identities using standard techniques (common denominator, Pythagorean identity, difference of squares). Part (ii) is a straightforward substitution followed by solving a linear equation in sin(2x) and finding solutions in the given range. Both parts are routine applications of AS-level techniques with no novel insight required, making this 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 | Marks | Guidance |
| \(\text{LHS} = \left(\frac{1}{c} - \frac{s}{c}\right)^2 = \frac{(1-s)(1-s)}{c^2} = \frac{(1-s)(1-s)}{1-s^2}\) | B1 | Expresses tan in terms of sin and cos |
| B1 | Correctly \(1 - s^2\) as the denominator | |
| \(= \frac{(1-s)(1-s)}{(1-s)(1+s)}\) | M1 | Factors and correct cancelling |
| \(\frac{1 - \sin x}{1 + \sin x}\) | A1 (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses part (i) to obtain \(\frac{1 - \sin 2x}{1 + \sin 2x} = \frac{1}{3} \rightarrow \sin 2x = \frac{1}{2}\) | M1 | Realises use of \(2x\) and makes \(\sin 2x\) the subject |
| \(x = \frac{\pi}{12}\) | A1 | Allow decimal \((0.262)\) |
| (or) \(x = \frac{5\pi}{12}\) | A1 | FT for \(\frac{\pi}{2} -\) 1st answer. Allow decimal \((1.31)\). \(\frac{\pi}{12}\) and \(\frac{5\pi}{12}\) only, and no others in range. SC \(\sin x = \frac{1}{2} \rightarrow \frac{\pi}{6}\ \frac{5\pi}{6}\) B1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{LHS} = \left(\frac{1}{c} - \frac{s}{c}\right)^2 = \frac{(1-s)(1-s)}{c^2} = \frac{(1-s)(1-s)}{1-s^2}$ | B1 | Expresses tan in terms of sin and cos |
| | B1 | Correctly $1 - s^2$ as the denominator |
| $= \frac{(1-s)(1-s)}{(1-s)(1+s)}$ | M1 | Factors and correct cancelling |
| $\frac{1 - \sin x}{1 + \sin x}$ | A1 (AG) | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses part (i) to obtain $\frac{1 - \sin 2x}{1 + \sin 2x} = \frac{1}{3} \rightarrow \sin 2x = \frac{1}{2}$ | M1 | Realises use of $2x$ and makes $\sin 2x$ the subject |
| $x = \frac{\pi}{12}$ | A1 | Allow decimal $(0.262)$ |
| (or) $x = \frac{5\pi}{12}$ | A1 | FT for $\frac{\pi}{2} -$ 1st answer. Allow decimal $(1.31)$. $\frac{\pi}{12}$ and $\frac{5\pi}{12}$ only, and no others in range. SC $\sin x = \frac{1}{2} \rightarrow \frac{\pi}{6}\ \frac{5\pi}{6}$ B1 |
---6 (i) Prove the identity $\left( \frac { 1 } { \cos x } - \tan x \right) ^ { 2 } \equiv \frac { 1 - \sin x } { 1 + \sin x }$.\\
(ii) Hence solve the equation $\left( \frac { 1 } { \cos 2 x } - \tan 2 x \right) ^ { 2 } = \frac { 1 } { 3 }$ for $0 \leqslant x \leqslant \pi$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q6 [7]}}