| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Convert sin/cos ratio to tan |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric equation question slightly above average difficulty. Part (i) requires dividing to get tan(2x-15°)=3, then solving with compound angle awareness. Part (ii) uses the standard identity sin²θ = 1-cos²θ to form a quadratic in cosθ. Both are textbook techniques with straightforward application, though the range restrictions and multi-step nature elevate it slightly above a routine exercise. |
| 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/Working | Mark | Guidance |
| \(3\sin(2x-15°) = \cos(2x-15°) \Rightarrow \tan(2x-15°) = \frac{1}{3}\) | M1 | Uses \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and reaches \(\tan(2x-15°) = \ldots\) |
| \(x = \frac{\tan^{-1}\left(\frac{1}{3}\right) \pm 15°}{2}\) | M1 | Correct strategy for finding \(x\) |
| One of \(x = 16.7°\) or \(-73.3°\) | A1 | One of awrt 16.7 or \(-73.3\) |
| \(x = 16.7°, -73.3°\) | A1 | Awrt 16.7 and awrt \(-73.3\) and no extras in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4\sin^2\theta + 8\cos\theta = 3 \Rightarrow 4(1-\cos^2\theta)+8\cos\theta=3 \Rightarrow 4\cos^2\theta - 8\cos\theta - 1 = 0\) | M1 | Applies \(\sin^2\theta = 1-\cos^2\theta\) and collects terms to obtain a 3TQ in \(\cos\theta\) |
| \(\cos\theta = \frac{8\pm\sqrt{64+4\times4}}{2\times4} \Rightarrow \theta = \cos^{-1}\left(1-\frac{\sqrt{5}}{2}\right) = \ldots\) | M1 | Solves their 3TQ and takes inverse cos to obtain at least one value for \(\theta\) |
| One of \(\theta = 1.69\) or \(4.59\) | A1 | Awrt 1.69 or 4.59 |
| \(\theta = 1.69, 4.59\) | A1 | Awrt 1.69 and awrt 4.59 and no extras in range |
## Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\sin(2x-15°) = \cos(2x-15°) \Rightarrow \tan(2x-15°) = \frac{1}{3}$ | M1 | Uses $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and reaches $\tan(2x-15°) = \ldots$ |
| $x = \frac{\tan^{-1}\left(\frac{1}{3}\right) \pm 15°}{2}$ | M1 | Correct strategy for finding $x$ |
| One of $x = 16.7°$ or $-73.3°$ | A1 | One of awrt 16.7 or $-73.3$ |
| $x = 16.7°, -73.3°$ | A1 | Awrt 16.7 and awrt $-73.3$ and no extras in range |
## Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\sin^2\theta + 8\cos\theta = 3 \Rightarrow 4(1-\cos^2\theta)+8\cos\theta=3 \Rightarrow 4\cos^2\theta - 8\cos\theta - 1 = 0$ | M1 | Applies $\sin^2\theta = 1-\cos^2\theta$ and collects terms to obtain a 3TQ in $\cos\theta$ |
| $\cos\theta = \frac{8\pm\sqrt{64+4\times4}}{2\times4} \Rightarrow \theta = \cos^{-1}\left(1-\frac{\sqrt{5}}{2}\right) = \ldots$ | M1 | Solves their 3TQ and takes inverse cos to obtain at least one value for $\theta$ |
| One of $\theta = 1.69$ or $4.59$ | A1 | Awrt 1.69 or 4.59 |
| $\theta = 1.69, 4.59$ | A1 | Awrt 1.69 and awrt 4.59 and no extras in range |7. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve, for $- 90 ^ { \circ } < x < 90 ^ { \circ }$, the equation
$$3 \sin \left( 2 x - 15 ^ { \circ } \right) = \cos \left( 2 x - 15 ^ { \circ } \right)$$
giving your answers to one decimal place.\\
(ii) Solve, for $0 < \theta < 2 \pi$, the equation
$$4 \sin ^ { 2 } \theta + 8 \cos \theta = 3$$
giving your answers to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel P2 2022 Q7 [8]}}