| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Identify student error in trig solution |
| Difficulty | Standard +0.3 This is a standard AS-level trigonometric equations question with routine techniques: converting cos²θ using Pythagorean identity to form a quadratic in sinθ, identifying common student errors (dividing by sinx=0 and missing solutions), and finding multiple solutions using periodicity. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses \(\cos^2\theta = 1 - \sin^2\theta\); \(5\cos^2\theta = 6\sin\theta \Rightarrow 5\sin^2\theta + 6\sin\theta - 5 = 0\) | M1, A1 | Forms 3TQ in \(\sin\theta\) |
| \(\Rightarrow \sin\theta = \frac{-3+\sqrt{34}}{5} \Rightarrow \theta = \ldots\) | dM1 | Solves 3TQ to produce one value of \(\theta\); dependent on having used \(\cos^2\theta = \pm1 \pm\sin^2\theta\) |
| \(\Rightarrow \theta = 34.5°, 145.5°, 394.5°\) | A1, A1 | Two of three correct awrt; all three correct and no other values |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| They cancel by \(\sin x\) and hence miss \(\sin x = 0 \Rightarrow x = 0\) | B1 | |
| They do not find all solutions of \(\cos x = \frac{3}{5}\) or miss \(x = -53.1°\) | B1 | Both errors required for full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Sets \(5\alpha + 40° = 720° - 53.1°\) | M1 | Sets \(5\alpha + 40° = 666.9°\) o.e. |
| \(\alpha = 125°\) | A1 | awrt \(\alpha = 125°\) |
## Question 12:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses $\cos^2\theta = 1 - \sin^2\theta$; $5\cos^2\theta = 6\sin\theta \Rightarrow 5\sin^2\theta + 6\sin\theta - 5 = 0$ | M1, A1 | Forms 3TQ in $\sin\theta$ |
| $\Rightarrow \sin\theta = \frac{-3+\sqrt{34}}{5} \Rightarrow \theta = \ldots$ | dM1 | Solves 3TQ to produce one value of $\theta$; dependent on having used $\cos^2\theta = \pm1 \pm\sin^2\theta$ |
| $\Rightarrow \theta = 34.5°, 145.5°, 394.5°$ | A1, A1 | Two of three correct awrt; all three correct and no other values |
### Part (ii)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| They cancel by $\sin x$ and hence miss $\sin x = 0 \Rightarrow x = 0$ | B1 | |
| They do not find all solutions of $\cos x = \frac{3}{5}$ or miss $x = -53.1°$ | B1 | Both errors required for full marks |
### Part (ii)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sets $5\alpha + 40° = 720° - 53.1°$ | M1 | Sets $5\alpha + 40° = 666.9°$ o.e. |
| $\alpha = 125°$ | A1 | awrt $\alpha = 125°$ |\begin{enumerate}
\item In this question you should show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
(i) Solve, for $0 < \theta \leqslant 450 ^ { \circ }$, the equation
$$5 \cos ^ { 2 } \theta = 6 \sin \theta$$
giving your answers to one decimal place.\\
(ii) (a) A student's attempt to solve the question\\
"Solve, for $- 90 ^ { \circ } < x < 90 ^ { \circ }$, the equation $3 \tan x - 5 \sin x = 0$ " is set out below.
$$\begin{gathered}
3 \tan x - 5 \sin x = 0 \\
3 \frac { \sin x } { \cos x } - 5 \sin x = 0 \\
3 \sin x - 5 \sin x \cos x = 0 \\
3 - 5 \cos x = 0 \\
\cos x = \frac { 3 } { 5 } \\
x = 53.1 ^ { \circ }
\end{gathered}$$
Identify two errors or omissions made by this student, giving a brief explanation of each.
The first four positive solutions, in order of size, of the equation
$$\cos \left( 5 \alpha + 40 ^ { \circ } \right) = \frac { 3 } { 5 }$$
are $\alpha _ { 1 } , \alpha _ { 2 } , \alpha _ { 3 }$ and $\alpha _ { 4 }$\\
(b) Find, to the nearest degree, the value of $\alpha _ { 4 }$
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q12 [9]}}