| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Mixed sin and cos linear |
| Difficulty | Standard +0.8 This is a Further Maths question requiring the t-substitution (half-angle substitution), which is a specialized technique beyond standard A-level. Part (a) involves algebraic manipulation with the substitution formulas, and part (b) requires solving the quadratic then converting back to find x values in the given range. While methodical once the technique is known, it's more demanding than typical A-level trig equations due to the Further Maths content and multi-step nature. |
| 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 |
| \(3\cos x - 2\sin x \Rightarrow 3\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{2t}{1+t^2}\right)\) | M1 | Uses at least one of \(\sin x = \frac{2t}{1+t^2}\) or \(\cos x = \frac{1-t^2}{1+t^2}\) to express \(3\cos x - 2\sin x\) in terms of \(t\) only |
| \(3\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{2t}{1+t^2}\right) = 1 \Rightarrow 3(1-t^2) - 4t = 1+t^2\) | M1 | Uses both correct formulae, equates to 1 and multiplies through by \(1+t^2\) to achieve quadratic in \(t\) |
| \(2t^2 + 2t - 1 = 0\) * | A1* | Collects terms to one side and simplifies to printed answer, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2 \cdot 2} \left\{= \frac{-1 \pm \sqrt{3}}{2} = -1.366..., 0.366...\right\}\) | M1 | Selects correct process to solve \(2t^2 + 2t - 1 = 0\); attempts by factorisation score M0 |
| \(\frac{x}{2} = \arctan(-1.366...)\) or \(\frac{x}{2} = \arctan(0.366...)\) and \(\Rightarrow x = ...\) | dM1 | Takes arctan of their \(t\) value and multiplies by 2; range \(-180° < x < 180°\) |
| \(x = \text{awrt } -107.6°\) or \(\text{awrt } 40.2°\) | A1 | One correct answer awrt 1 d.p. |
| \(x = -107.6°,\ 40.2°\) cao | A1 | Both correct answers to 1 d.p., no incorrect answers in range \(-180° < x < 180°\) |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\cos x - 2\sin x \Rightarrow 3\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{2t}{1+t^2}\right)$ | M1 | Uses at least one of $\sin x = \frac{2t}{1+t^2}$ or $\cos x = \frac{1-t^2}{1+t^2}$ to express $3\cos x - 2\sin x$ in terms of $t$ only |
| $3\left(\frac{1-t^2}{1+t^2}\right) - 2\left(\frac{2t}{1+t^2}\right) = 1 \Rightarrow 3(1-t^2) - 4t = 1+t^2$ | M1 | Uses both correct formulae, equates to 1 and multiplies through by $1+t^2$ to achieve quadratic in $t$ |
| $2t^2 + 2t - 1 = 0$ * | A1* | Collects terms to one side and simplifies to printed answer, no errors seen |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2 \cdot 2} \left\{= \frac{-1 \pm \sqrt{3}}{2} = -1.366..., 0.366...\right\}$ | M1 | Selects correct process to solve $2t^2 + 2t - 1 = 0$; attempts by factorisation score M0 |
| $\frac{x}{2} = \arctan(-1.366...)$ or $\frac{x}{2} = \arctan(0.366...)$ and $\Rightarrow x = ...$ | dM1 | Takes arctan of their $t$ value and multiplies by 2; range $-180° < x < 180°$ |
| $x = \text{awrt } -107.6°$ or $\text{awrt } 40.2°$ | A1 | One correct answer awrt 1 d.p. |
| $x = -107.6°,\ 40.2°$ cao | A1 | Both correct answers to 1 d.p., no incorrect answers in range $-180° < x < 180°$ |
---\begin{enumerate}
\item (a) Use the substitution $t = \tan \left( \frac { x } { 2 } \right)$ to show that the equation
\end{enumerate}
$$3 \cos x - 2 \sin x = 1$$
can be written in the form
$$2 t ^ { 2 } + 2 t - 1 = 0$$
(b) Hence solve, for $- 180 ^ { \circ } < x < 180 ^ { \circ }$, the equation
$$3 \cos x - 2 \sin x = 1$$
giving your answers to one decimal place.
\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q2 [7]}}