| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) is a standard trigonometric identity proof requiring manipulation of sec and tan into a single fraction, then using the Pythagorean identity—routine for AS level. Part (b) applies the proven identity to form a quadratic in sin(2x), requiring double angle work and inverse trig, but follows a clear path once the identity is recognized. This is slightly easier than average due to the scaffolded structure and standard techniques. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05j 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{\cos\theta}+\tan\theta = \frac{1+\sin\theta}{\cos\theta}\) or \(\frac{(1+\sin\theta)\cos\theta}{\cos^2\theta}\) | M1 | 1.1b — Combines terms with common denominator; numerator must be correct for their denominator. Condone different variable for \(\theta\) except for final mark |
| \(=\frac{1+\sin\theta}{\cos\theta}\times\frac{1-\sin\theta}{1-\sin\theta}=\frac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)}=\frac{\cos^2\theta}{\cos\theta(1-\sin\theta)}\) or \(\frac{(1+\sin\theta)\cos\theta}{\cos^2\theta}=\frac{(1+\sin\theta)\cos\theta}{1-\sin^2\theta}=\frac{(1+\sin\theta)\cos\theta}{(1+\sin\theta)(1-\sin\theta)}\) | dM1 | 2.1 — Dependent on M1. Either: multiply by \(\frac{1-\sin\theta}{1-\sin\theta}\), use difference of squares and \(\cos^2\theta=1-\sin^2\theta\); or use \(\cos^2\theta=1-\sin^2\theta\) on denominator and difference of squares |
| \(=\frac{\cos\theta}{1-\sin\theta}\) | A1* | 1.1b — Fully correct proof, correct notation, no errors. Withhold for e.g. writing \(\sin\) instead of \(\sin\theta\), or \(\sin\theta^2\) instead of \(\sin^2\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{\cos 2x}+\tan 2x=3\cos 2x \Rightarrow 1+\sin 2x=3\cos^2 2x=3(1-\sin^2 2x)\) or \(\frac{\cos 2x}{1-\sin 2x}=3\cos 2x \Rightarrow \cos 2x(2-3\sin 2x)=0\) | M1 | 2.1 — Multiplies through by \(\cos 2x\) and applies \(\cos^2 2x=1-\sin^2 2x\), or multiplies \(1-\sin 2x\) and reaches equation in \(\cos 2x\) and \(\sin 2x\) |
| \(\Rightarrow 3\sin^2 2x+\sin 2x-2=0\) or \(\Rightarrow \cos 2x(2-3\sin 2x)=0\) | A1 | 1.1b — Correct equation; \(=0\) may be implied by later work. Condone notational slips |
| \(\sin 2x=\frac{2}{3},\ (-1) \Rightarrow 2x=...\Rightarrow x=...\) | M1 | 1.1b — Solves for \(\sin 2x\), uses arcsin for at least one value of \(2x\), divides by 2. Roots of quadratic may be found by calculator |
| \(x=20.9°\) | A1 | 1.1b — For 1 required angle; accept awrt 21 or awrt 69, or awrt 0.36 rad or awrt 1.21 rad |
| \(x=69.1°\) (and no others in range) | A1 | 1.1b — Both angles (awrt 20.9 and awrt 69.1) and no others in range. If \(x=45°\) found it must be rejected. Condone notational slips |
# Question 13:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{\cos\theta}+\tan\theta = \frac{1+\sin\theta}{\cos\theta}$ or $\frac{(1+\sin\theta)\cos\theta}{\cos^2\theta}$ | M1 | 1.1b — Combines terms with common denominator; numerator must be correct for their denominator. Condone different variable for $\theta$ except for final mark |
| $=\frac{1+\sin\theta}{\cos\theta}\times\frac{1-\sin\theta}{1-\sin\theta}=\frac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)}=\frac{\cos^2\theta}{\cos\theta(1-\sin\theta)}$ **or** $\frac{(1+\sin\theta)\cos\theta}{\cos^2\theta}=\frac{(1+\sin\theta)\cos\theta}{1-\sin^2\theta}=\frac{(1+\sin\theta)\cos\theta}{(1+\sin\theta)(1-\sin\theta)}$ | dM1 | 2.1 — Dependent on M1. Either: multiply by $\frac{1-\sin\theta}{1-\sin\theta}$, use difference of squares and $\cos^2\theta=1-\sin^2\theta$; **or** use $\cos^2\theta=1-\sin^2\theta$ on denominator and difference of squares |
| $=\frac{\cos\theta}{1-\sin\theta}$ | A1* | 1.1b — Fully correct proof, correct notation, no errors. Withhold for e.g. writing $\sin$ instead of $\sin\theta$, or $\sin\theta^2$ instead of $\sin^2\theta$ |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{\cos 2x}+\tan 2x=3\cos 2x \Rightarrow 1+\sin 2x=3\cos^2 2x=3(1-\sin^2 2x)$ **or** $\frac{\cos 2x}{1-\sin 2x}=3\cos 2x \Rightarrow \cos 2x(2-3\sin 2x)=0$ | M1 | 2.1 — Multiplies through by $\cos 2x$ and applies $\cos^2 2x=1-\sin^2 2x$, or multiplies $1-\sin 2x$ and reaches equation in $\cos 2x$ and $\sin 2x$ |
| $\Rightarrow 3\sin^2 2x+\sin 2x-2=0$ **or** $\Rightarrow \cos 2x(2-3\sin 2x)=0$ | A1 | 1.1b — Correct equation; $=0$ may be implied by later work. Condone notational slips |
| $\sin 2x=\frac{2}{3},\ (-1) \Rightarrow 2x=...\Rightarrow x=...$ | M1 | 1.1b — Solves for $\sin 2x$, uses arcsin for at least one value of $2x$, divides by 2. Roots of quadratic may be found by calculator |
| $x=20.9°$ | A1 | 1.1b — For 1 required angle; accept awrt 21 or awrt 69, or awrt 0.36 rad or awrt 1.21 rad |
| $x=69.1°$ (and no others in range) | A1 | 1.1b — Both angles (awrt 20.9 and awrt 69.1) and no others in range. If $x=45°$ found it must be rejected. Condone notational slips |\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
$$\frac { 1 } { \cos \theta } + \tan \theta \equiv \frac { \cos \theta } { 1 - \sin \theta } \quad \theta \neq ( 2 n + 1 ) 90 ^ { \circ } \quad n \in \mathbb { Z }$$
Given that $\cos 2 x \neq 0$\\
(b) solve for $0 < x < 90 ^ { \circ }$
$$\frac { 1 } { \cos 2 x } + \tan 2 x = 3 \cos 2 x$$
giving your answers to one decimal place.
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q13 [8]}}