| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: tan/sin/cos identity manipulation |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard trigonometric manipulation (converting tan to sin/cos, using the Pythagorean identity) followed by solving a quadratic equation. The algebraic steps are routine for P1 level, and part (i) provides the target form, making it slightly easier than average. The only mild challenge is recognizing which solutions are valid in the given range. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4\cos^2\theta + 15\sin\theta = 0\) | M1 | Replace \(\tan\theta\) by \(\frac{\sin\theta}{\cos\theta}\) and multiply by \(\sin\theta\) or equivalent |
| \(4(1-s^2)+15s = 0 \rightarrow 4\sin^2\theta - 15\sin\theta - 4 = 0\) | M1A1 [3] | Use \(c^2 = 1-s^2\) and rearrange to AG (www) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\theta = -1/4\) | B1 | |
| \(\theta = 194.5\) or \(345.5\) | B1B1\(\checkmark\) [3] | Ignore other solution; ft from 1st solution; SC B1 both angles in rads (3.39 and 6.03) |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\cos^2\theta + 15\sin\theta = 0$ | M1 | Replace $\tan\theta$ by $\frac{\sin\theta}{\cos\theta}$ and multiply by $\sin\theta$ or equivalent |
| $4(1-s^2)+15s = 0 \rightarrow 4\sin^2\theta - 15\sin\theta - 4 = 0$ | M1A1 [3] | Use $c^2 = 1-s^2$ and rearrange to AG (www) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = -1/4$ | B1 | |
| $\theta = 194.5$ or $345.5$ | B1B1$\checkmark$ [3] | Ignore other solution; ft from 1st solution; SC B1 both angles in rads (3.39 and 6.03) |
---4 (i) Show that the equation $\frac { 4 \cos \theta } { \tan \theta } + 15 = 0$ can be expressed as
$$4 \sin ^ { 2 } \theta - 15 \sin \theta - 4 = 0$$
(ii) Hence solve the equation $\frac { 4 \cos \theta } { \tan \theta } + 15 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2015 Q4 [6]}}