| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.8 Part (a) is a standard reciprocal identity proof requiring manipulation of cosec and cot definitions. Part (b) is more challenging: after using the identity, students must recognize that cos x cot(3x-50°) = cos x cot x implies cot(3x-50°) = cot x, leading to angle equations 3x-50° = x + 180n°. This requires understanding cotangent periodicity and solving for multiple solutions in the given range, which goes beyond routine exercises. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\csc\theta = \frac{1}{\sin\theta}\) | B1 | Do not accept \(\csc\theta = \frac{1}{\sin}\) with \(\theta\) missing |
| \(\csc\theta - \sin\theta = \frac{1}{\sin\theta} - \sin\theta = \frac{1-\sin^2\theta}{\sin\theta}\) | M1 | Key step: forming a single fraction/common denominator |
| \(= \frac{\cos^2\theta}{\sin\theta} = \cos\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta\cot\theta\) | A1* | Shows careful work with all necessary steps. No notational or bracketing errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\cot\theta = \frac{\cos\theta}{\sin\theta}\) | B1 | |
| \(\cos\theta\cot\theta = \frac{\cos^2\theta}{\sin\theta} = \frac{1-\sin^2\theta}{\sin\theta}\) | M1 | |
| \(= \frac{1}{\sin\theta} - \sin\theta = \csc\theta - \sin\theta\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\csc x - \sin x = \cos x\cot(3x-50°)\) \(\Rightarrow \cos x\cot x = \cos x\cot(3x-50°)\) | Uses part (a), cancels/factorises \(\cos x\) term | |
| \(\cot x = \cot(3x-50°) \Rightarrow x = 3x - 50°\) | M1 | |
| \(x = 25°\) | A1 | |
| Also \(\cot x = \cot(3x-50°) \Rightarrow x + 180° = 3x - 50°\) | M1 | Key step: \(\cot x\) has period \(180°\), finding second solution |
| \(x = 115°\) | A1 | Withhold if additional values in range \((0°, 180°)\) |
| Deduces \(x = 90°\) | B1 | From \(\cos x = 0 \Rightarrow x = 90°\). Ignore additional values |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\csc\theta = \frac{1}{\sin\theta}$ | B1 | Do not accept $\csc\theta = \frac{1}{\sin}$ with $\theta$ missing |
| $\csc\theta - \sin\theta = \frac{1}{\sin\theta} - \sin\theta = \frac{1-\sin^2\theta}{\sin\theta}$ | M1 | Key step: forming a single fraction/common denominator |
| $= \frac{\cos^2\theta}{\sin\theta} = \cos\theta \times \frac{\cos\theta}{\sin\theta} = \cos\theta\cot\theta$ | A1* | Shows careful work with all necessary steps. No notational or bracketing errors |
**Alt 1 (RHS to LHS):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\cot\theta = \frac{\cos\theta}{\sin\theta}$ | B1 | |
| $\cos\theta\cot\theta = \frac{\cos^2\theta}{\sin\theta} = \frac{1-\sin^2\theta}{\sin\theta}$ | M1 | |
| $= \frac{1}{\sin\theta} - \sin\theta = \csc\theta - \sin\theta$ | A1* | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\csc x - \sin x = \cos x\cot(3x-50°)$ $\Rightarrow \cos x\cot x = \cos x\cot(3x-50°)$ | | Uses part (a), cancels/factorises $\cos x$ term |
| $\cot x = \cot(3x-50°) \Rightarrow x = 3x - 50°$ | M1 | |
| $x = 25°$ | A1 | |
| Also $\cot x = \cot(3x-50°) \Rightarrow x + 180° = 3x - 50°$ | M1 | Key step: $\cot x$ has period $180°$, finding second solution |
| $x = 115°$ | A1 | Withhold if additional values in range $(0°, 180°)$ |
| Deduces $x = 90°$ | B1 | From $\cos x = 0 \Rightarrow x = 90°$. Ignore additional values |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that
$$\operatorname { cosec } \theta - \sin \theta \equiv \cos \theta \cot \theta \quad \theta \neq ( 180 n ) ^ { \circ } \quad n \in \mathbb { Z }$$
(b) Hence, or otherwise, solve for $0 < x < 180 ^ { \circ }$
$$\operatorname { cosec } x - \sin x = \cos x \cot \left( 3 x - 50 ^ { \circ } \right)$$
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q12 [8]}}