| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Multiple angle equations |
| Difficulty | Standard +0.8 This question requires multiple sophisticated steps: converting tan and sec to sin/cos, manipulating to form a quadratic in sin x, then applying a compound angle substitution and carefully handling the range restrictions. The 'show that' proof in part (i) requires algebraic insight beyond routine manipulation, and part (ii) demands careful attention to the transformed domain when solving for x from the compound angle. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\tan x + 3\cos x + \frac{1}{\cos x} = 0 \rightarrow 4\sin x + 3\cos^2 x + 1 = 0\) | M1 | Multiply by \(\cos x\) or common denominator of \(\cos x\) |
| \(4\sin x + 3(1-\sin^2 x) + 1 = 0 \rightarrow 3\sin^2 x - 4\sin x - 4 = 0\) | M1 | Use \(\cos^2 x = 1 - \sin^2 x\) and simplify to 3-term quadratic in \(\sin x\) |
| \(\sin x = -\frac{2}{3}\) | A1 | AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x - 20° = 221.8°, 318.2°\) | M1A1 | Attempt to solve \(\sin(2x-20) = -2/3\) (M1). At least 1 correct (A1) |
| \(x = 120.9°, 169.1°\) | A1, A1FT | FT for \(290°\) - other solution. SC A1 both answers in radians |
| 4 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\tan x + 3\cos x + \frac{1}{\cos x} = 0 \rightarrow 4\sin x + 3\cos^2 x + 1 = 0$ | M1 | Multiply by $\cos x$ or common denominator of $\cos x$ |
| $4\sin x + 3(1-\sin^2 x) + 1 = 0 \rightarrow 3\sin^2 x - 4\sin x - 4 = 0$ | M1 | Use $\cos^2 x = 1 - \sin^2 x$ and simplify to 3-term quadratic in $\sin x$ |
| $\sin x = -\frac{2}{3}$ | A1 | AG |
| | **3** | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x - 20° = 221.8°, 318.2°$ | M1A1 | Attempt to solve $\sin(2x-20) = -2/3$ (M1). At least 1 correct (A1) |
| $x = 120.9°, 169.1°$ | A1, A1FT | FT for $290°$ - other solution. SC A1 both answers in radians |
| | **4** | |5 (i) Given that $4 \tan x + 3 \cos x + \frac { 1 } { \cos x } = 0$, show, without using a calculator, that $\sin x = - \frac { 2 } { 3 }$.\\
(ii) Hence, showing all necessary working, solve the equation
$$4 \tan \left( 2 x - 20 ^ { \circ } \right) + 3 \cos \left( 2 x - 20 ^ { \circ } \right) + \frac { 1 } { \cos \left( 2 x - 20 ^ { \circ } \right) } = 0$$
for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q5 [7]}}