| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Standard +0.3 This is a standard P2 trigonometric equations question requiring routine techniques: (i) involves taking square roots, using inverse tan, and adjusting for the compound angle; (ii) requires expanding, using the Pythagorean identity to form a quadratic in sin θ or cos θ, then solving. Both parts are multi-step but follow well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan^2\!\left(2x+\frac{\pi}{4}\right)=3 \Rightarrow 2x+\frac{\pi}{4}=\pm\frac{\pi}{3}, \pm\frac{2\pi}{3}\) | M1 | Correct order of operations: square root then arctan. Condone using \(\theta\) for \(2x+\frac{\pi}{4}\) |
| \(\Rightarrow x = \left(\ldots - \frac{\pi}{4}\right) \div 2\) | dM1 | Complete attempt to find one value for \(x\), moving \(\frac{\pi}{4}\) then dividing by 2 |
| \(\Rightarrow x = \frac{\pi}{24},\ -\frac{11\pi}{24},\ \frac{5\pi}{24},\ -\frac{7\pi}{24}\) | A1; A1, A1 | First A1 for one of \(\frac{\pi}{24}\) or \(-\frac{11\pi}{24}\); second A1 for one of \(\frac{5\pi}{24}\) or \(-\frac{7\pi}{24}\); third A1 for all four and no extras in range |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((2\sin\theta - \cos\theta)^2 = 1 \Rightarrow 4\sin^2\theta - 4\sin\theta\cos\theta + \cos^2\theta = 1\); uses \(\sin^2\theta+\cos^2\theta=1 \Rightarrow 4\sin^2\theta - 4\sin\theta\cos\theta = \sin^2\theta\) | M1 | Attempts to multiply out to at least three terms and uses \(\sin^2\theta+\cos^2\theta=1\) somewhere in equation |
| \(\Rightarrow \sin\theta(3\sin\theta - 4\cos\theta) = 0\) | dM1 | Cancels or factorises out the \(\sin\theta\) term to produce a factor \(a\sin\theta \pm b\cos\theta\) or equation of form \(a\sin\theta \pm b\cos\theta = 0\) |
| \(\tan\theta = \frac{4}{3}\) | A1 | |
| \(\theta = 53.1°,\ 233.1°\) | A1 | No others in range; note 0° and 360° are outside range |
| \(\theta = 180°\) | B1 | Award when seen; allow however it arises |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((2\sin\theta-\cos\theta)^2=1 \Rightarrow 2\sin\theta-\cos\theta=\pm1\); rearranges and squares; uses \(\sin^2\theta+\cos^2\theta=1 \Rightarrow 5\cos^2\theta \pm 2\cos\theta - 3 = 0\) | M1 | Attempts at least one of \(2\sin\theta-\cos\theta=\pm1\), rearranges, squares, uses identity |
| Solves quadratic in \(\cos\theta\) or \(\sin\theta\) | dM1 | To obtain at least one value for \(\cos\theta\) or \(\sin\theta\) |
| \(\cos\theta = \frac{3}{5},\ -1\) or \(\cos\theta = -\frac{3}{5},\ 1\) | A1 | One correct pair of solutions |
| \(\theta = 53.1°,\ 233.1°\) | A1 | No others in range |
| \(\theta = 180°\) | B1 | Award when seen; allow however it arises |
## Question 10(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan^2\!\left(2x+\frac{\pi}{4}\right)=3 \Rightarrow 2x+\frac{\pi}{4}=\pm\frac{\pi}{3}, \pm\frac{2\pi}{3}$ | M1 | Correct order of operations: square root then arctan. Condone using $\theta$ for $2x+\frac{\pi}{4}$ |
| $\Rightarrow x = \left(\ldots - \frac{\pi}{4}\right) \div 2$ | dM1 | Complete attempt to find one value for $x$, moving $\frac{\pi}{4}$ then dividing by 2 |
| $\Rightarrow x = \frac{\pi}{24},\ -\frac{11\pi}{24},\ \frac{5\pi}{24},\ -\frac{7\pi}{24}$ | A1; A1, A1 | First A1 for one of $\frac{\pi}{24}$ or $-\frac{11\pi}{24}$; second A1 for one of $\frac{5\pi}{24}$ or $-\frac{7\pi}{24}$; third A1 for all four and no extras in range |
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## Question 10(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $(2\sin\theta - \cos\theta)^2 = 1 \Rightarrow 4\sin^2\theta - 4\sin\theta\cos\theta + \cos^2\theta = 1$; uses $\sin^2\theta+\cos^2\theta=1 \Rightarrow 4\sin^2\theta - 4\sin\theta\cos\theta = \sin^2\theta$ | M1 | Attempts to multiply out to at least three terms and uses $\sin^2\theta+\cos^2\theta=1$ somewhere in equation |
| $\Rightarrow \sin\theta(3\sin\theta - 4\cos\theta) = 0$ | dM1 | Cancels or factorises out the $\sin\theta$ term to produce a factor $a\sin\theta \pm b\cos\theta$ or equation of form $a\sin\theta \pm b\cos\theta = 0$ |
| $\tan\theta = \frac{4}{3}$ | A1 | |
| $\theta = 53.1°,\ 233.1°$ | A1 | No others in range; note 0° and 360° are outside range |
| $\theta = 180°$ | B1 | Award when seen; allow however it arises |
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## Question 10(ii) Alt:
| Working | Mark | Guidance |
|---------|------|----------|
| $(2\sin\theta-\cos\theta)^2=1 \Rightarrow 2\sin\theta-\cos\theta=\pm1$; rearranges and squares; uses $\sin^2\theta+\cos^2\theta=1 \Rightarrow 5\cos^2\theta \pm 2\cos\theta - 3 = 0$ | M1 | Attempts at least one of $2\sin\theta-\cos\theta=\pm1$, rearranges, squares, uses identity |
| Solves quadratic in $\cos\theta$ or $\sin\theta$ | dM1 | To obtain at least one value for $\cos\theta$ or $\sin\theta$ |
| $\cos\theta = \frac{3}{5},\ -1$ or $\cos\theta = -\frac{3}{5},\ 1$ | A1 | One correct pair of solutions |
| $\theta = 53.1°,\ 233.1°$ | A1 | No others in range |
| $\theta = 180°$ | B1 | Award when seen; allow however it arises |10. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve, for $- \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$
$$\tan ^ { 2 } \left( 2 x + \frac { \pi } { 4 } \right) = 3$$
(ii) Solve, for $0 < \theta < 360 ^ { \circ }$
$$( 2 \sin \theta - \cos \theta ) ^ { 2 } = 1$$
giving your answers, as appropriate, to one decimal place.\\
\hfill \mbox{\textit{Edexcel P2 2021 Q10 [10]}}