| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.8 This is a two-part question requiring proof of a non-trivial trigonometric identity using double angle formulas and reciprocal functions, followed by applying this result to solve a more complex equation. Part (a) requires algebraic manipulation with sin 2x and cos 2x expansions, while part (b) demands recognizing the pattern from part (a) with substitution (x → 2θ) and then solving a resulting trigonometric equation. The multi-step nature, need for strategic substitution, and combination of multiple trigonometric concepts places this above average difficulty but within reach of well-prepared P3 students. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(\sin 2x = 2\sin x\cos x\) AND \(\cos 2x = 1-2\sin^2 x\) in \(\frac{\sin 2x}{\cos x}+\frac{\cos 2x}{\sin x}\) | M1 | Condone sign slips on versions of \(\cos 2x\) |
| \(\frac{\sin 2x}{\cos x}+\frac{\cos 2x}{\sin x} = \frac{2\sin x\cos x}{\cos x}+\frac{1-2\sin^2 x}{\sin x} = \frac{2\sin x\cancel{\cos x}}{\cancel{\cos x}}+\frac{1}{\sin x}-\frac{2\sin^2 x}{\sin x} = \frac{1}{\sin x} = \cosec x\) | dM1 | Adopts a valid approach that can be followed and completes the proof. Condone writing \(\cos\) for \(\cos x\) or \(\sin x^2\) for \(\sin^2 x\) |
| Correct proof showing all necessary intermediate steps with no errors | A1* | LHS starting point does not need to be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses part (a) \(\Rightarrow 7+\cosec 2\theta = 3\cot^2 2\theta\) | B1 | Watch for and do not allow \(7+\cosec\mathbf{2\theta}3\cot^2\) |
| Uses \(\pm1\pm\cot^2 2\theta = \pm\cosec^2 2\theta \rightarrow\) 3TQ in \(\cosec 2\theta\); OR replaces \(\cosec 2\theta\) with \(\frac{1}{\sin 2\theta}\), \(\cot^2 2\theta\) with \(\frac{\cos^2 2\theta}{\sin^2 2\theta}\), multiplies by \(\sin^2 2\theta\) and uses \(\pm\cos^2 2\theta = \pm1\pm\sin^2 2\theta \rightarrow\) 3TQ in \(\sin 2\theta\) | M1 | Condone \(3\cot^2 2\theta\) being replaced by \(3\times\pm\cosec^2 2\theta\pm1\); "7" missing scores max 2 marks |
| \(3\cosec^2 2\theta - \cosec 2\theta - 10 = 0\) or \(10\sin^2 2\theta + \sin 2\theta - 3 = 0\) | A1 | The \(=0\) may be implied by further work |
| \((3\cosec 2\theta+5)(\cosec 2\theta-2)=0\) or \((5\sin 2\theta+3)(2\sin 2\theta-1)=0\) \(\Rightarrow \cosec 2\theta = -\frac{5}{3}, 2\) or \(\Rightarrow \sin 2\theta = -\frac{3}{5}, \frac{1}{2}\) | dM1 | For a correct attempt to solve their 3TQ in \(\sin 2\theta\) or \(\cosec 2\theta\) leading to a value for \(\theta\). If they only state \(\sin\theta = -\frac{3}{5}, \frac{1}{2}\) without proceeding to arcsin and \(\div 2\), it is M0 |
| \(\theta = \frac{\pi}{12}(0.262),\ \frac{5\pi}{12}(1.31),\ -0.322,\ -1.25\) | A1, A1 | awrt these values; second A1 for all four values with no additional values within the range |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\sin 2x = 2\sin x\cos x$ AND $\cos 2x = 1-2\sin^2 x$ in $\frac{\sin 2x}{\cos x}+\frac{\cos 2x}{\sin x}$ | M1 | Condone sign slips on versions of $\cos 2x$ |
| $\frac{\sin 2x}{\cos x}+\frac{\cos 2x}{\sin x} = \frac{2\sin x\cos x}{\cos x}+\frac{1-2\sin^2 x}{\sin x} = \frac{2\sin x\cancel{\cos x}}{\cancel{\cos x}}+\frac{1}{\sin x}-\frac{2\sin^2 x}{\sin x} = \frac{1}{\sin x} = \cosec x$ | dM1 | Adopts a valid approach that can be followed and completes the proof. Condone writing $\cos$ for $\cos x$ or $\sin x^2$ for $\sin^2 x$ |
| Correct proof showing all necessary intermediate steps with no errors | A1* | LHS starting point does not need to be seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses part (a) $\Rightarrow 7+\cosec 2\theta = 3\cot^2 2\theta$ | B1 | Watch for and do not allow $7+\cosec\mathbf{2\theta}3\cot^2$ |
| Uses $\pm1\pm\cot^2 2\theta = \pm\cosec^2 2\theta \rightarrow$ 3TQ in $\cosec 2\theta$; OR replaces $\cosec 2\theta$ with $\frac{1}{\sin 2\theta}$, $\cot^2 2\theta$ with $\frac{\cos^2 2\theta}{\sin^2 2\theta}$, multiplies by $\sin^2 2\theta$ and uses $\pm\cos^2 2\theta = \pm1\pm\sin^2 2\theta \rightarrow$ 3TQ in $\sin 2\theta$ | M1 | Condone $3\cot^2 2\theta$ being replaced by $3\times\pm\cosec^2 2\theta\pm1$; "7" missing scores max 2 marks |
| $3\cosec^2 2\theta - \cosec 2\theta - 10 = 0$ **or** $10\sin^2 2\theta + \sin 2\theta - 3 = 0$ | A1 | The $=0$ may be implied by further work |
| $(3\cosec 2\theta+5)(\cosec 2\theta-2)=0$ **or** $(5\sin 2\theta+3)(2\sin 2\theta-1)=0$ $\Rightarrow \cosec 2\theta = -\frac{5}{3}, 2$ **or** $\Rightarrow \sin 2\theta = -\frac{3}{5}, \frac{1}{2}$ | dM1 | For a correct attempt to solve their 3TQ in $\sin 2\theta$ or $\cosec 2\theta$ leading to a value for $\theta$. If they only state $\sin\theta = -\frac{3}{5}, \frac{1}{2}$ without proceeding to arcsin and $\div 2$, it is M0 |
| $\theta = \frac{\pi}{12}(0.262),\ \frac{5\pi}{12}(1.31),\ -0.322,\ -1.25$ | A1, A1 | awrt these values; second A1 for all four values with no additional values within the range |
---7. (a) Prove that
$$\frac { \sin 2 x } { \cos x } + \frac { \cos 2 x } { \sin x } \equiv \operatorname { cosec } x \quad x \neq \frac { n \pi } { 2 } n \in \mathbb { Z }$$
(b) Hence solve, for $- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$
$$7 + \frac { \sin 4 \theta } { \cos 2 \theta } + \frac { \cos 4 \theta } { \sin 2 \theta } = 3 \cot ^ { 2 } 2 \theta$$
giving your answers in radians to 3 significant figures where appropriate.\\
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\hfill \mbox{\textit{Edexcel P3 2021 Q7 [9]}}