| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: algebraic fraction manipulation |
| Difficulty | Moderate -0.3 This is a standard A-level trigonometric equation requiring algebraic manipulation to convert to quadratic form (given in part i) then solving. The manipulation involves clearing fractions and using the Pythagorean identity, which are routine techniques. Part (ii) is straightforward once the quadratic is solved. Slightly easier than average due to the scaffolding in part (i) showing the target form. |
| 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 | Marks | Guidance |
| \(\cos\theta+4+5\sin^2\theta+5\sin\theta-5\sin\theta-5\ (=0)\) | M1 | Multiply throughout by \(\sin\theta+1\). Accept if \(5\sin\theta-5\sin\theta\) is not seen |
| \(5(1-\cos^2\theta)+\cos\theta-1\ (=0)\) | M1 | Use \(s^2=1-c^2\) |
| \(5\cos^2\theta-\cos\theta-4=0\) AG | A1 | Rearrange to AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta=1\) and \(-0.8\) | B1 | Both required |
| \(\theta=[0°, 360°],\ [143.1°],\ [216.9°]\) | B1 B1 B1 FT | Both solutions required for 1st mark. For 3rd mark FT for \((360°- their\ 143.1°)\). Extra solution(s) in range (e.g. 180°) among 4 correct solutions scores \(\frac{3}{4}\) |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta+4+5\sin^2\theta+5\sin\theta-5\sin\theta-5\ (=0)$ | M1 | Multiply throughout by $\sin\theta+1$. Accept if $5\sin\theta-5\sin\theta$ is not seen |
| $5(1-\cos^2\theta)+\cos\theta-1\ (=0)$ | M1 | Use $s^2=1-c^2$ |
| $5\cos^2\theta-\cos\theta-4=0$ AG | A1 | Rearrange to AG |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta=1$ and $-0.8$ | B1 | Both required |
| $\theta=[0°, 360°],\ [143.1°],\ [216.9°]$ | B1 B1 B1 FT | Both solutions required for 1st mark. For 3rd mark FT for $(360°- their\ 143.1°)$. Extra solution(s) in range (e.g. 180°) among 4 correct solutions scores $\frac{3}{4}$ |
---5 (i) Show that the equation $\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0$ may be expressed as $5 \cos ^ { 2 } \theta - \cos \theta - 4 = 0$.\\[0pt]
[3]\\
(ii) Hence solve the equation $\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q5 [7]}}