Edexcel FP1 AS 2024 June — Question 4 12 marks

Exam BoardEdexcel
ModuleFP1 AS (Further Pure 1 AS)
Year2024
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeMixed sin and cos linear
DifficultyStandard +0.3 This is a standard Further Pure 1 question on the t-substitution (half-angle substitution) for trigonometric equations. Part (a) is a routine proof of a standard identity, part (b) involves algebraic manipulation using given identities, and part (c) requires solving a factored equation. While it requires multiple steps and is from Further Maths, the techniques are well-practiced and follow a predictable pattern with no novel insight required, making it slightly easier than average overall.
Spec1.05j 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 intervals1.05p Proof involving trig: functions and identities

  1. (a) Given that \(t = \tan \frac { X } { 2 }\) prove that
$$\cos x \equiv \frac { 1 - t ^ { 2 } } { 1 + t ^ { 2 } }$$ (b) Show that the equation $$3 \tan x - 10 \cos x = 10$$ can be written in the form $$( t + 2 ) \left( a t ^ { 2 } + b t + c \right) = 0$$ where \(t = \tan \frac { X } { 2 }\) and \(a , b\) and \(c\) are integers to be determined.
(c) Hence solve, for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\), the equation $$3 \tan x - 10 \cos x = 10$$

Question 4(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}\) or \(2\cos^2\frac{x}{2}-1\) or \(1-2\sin^2\frac{x}{2}\)B1 Any correct appropriate double angle identity for \(\cos x\) or possibly \(\tan x\)
\(\Rightarrow \cos x = \left(\frac{1}{\sqrt{1+t^2}}\right)^2-\left(\frac{t}{\sqrt{1+t^2}}\right)^2\)M1 Substitutes correct expressions for \(\cos\frac{x}{2}\) and/or \(\sin\frac{x}{2}\) in terms of \(t\)
\(= \frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}\) *A1* Completes proof with sufficient working shown, no errors; given mark
Question 4 (Trigonometry with t-substitution):
Part (a):
AnswerMarks Guidance
WorkingMark Guidance
\(= \frac{2t}{1+t^2} \times \frac{1-t^2}{2t} = \frac{1-t^2}{1+t^2}\)M1 Substituting correct expressions for \(\cos\frac{x}{2}\), \(\sin\frac{x}{2}\) and \(\tan\frac{x}{2}\) in terms of \(t\)
Completion to \(\frac{1-t^2}{1+t^2}\)A1* Completing the proof with sufficient working shown, no errors. Allow \(\theta\) for \(x\) for first 2 marks but must obtain \(\cos x = \ldots\) for A1*
Special Case (quoting \(\tan x\) and/or \(\sin x\) to verify): Scores SC B1 only
Part (b):
AnswerMarks Guidance
WorkingMark Guidance
\(3\tan x - 10\cos x = 10 \Rightarrow 3 \times \frac{2t}{1-t^2} - 10 \times \frac{1-t^2}{1+t^2} = 10\)B1 Applies part (a) and half angle formula for \(\tan\) to give correct equation in \(t\)
\(\Rightarrow 6t(1+t^2) - 10(1-t^2)^2 = 10(1-t^2)(1+t^2)\) leading to \(6t^3 + 20t^2 + 6t - 20 = 0\) or \(3t^3 + 10t^2 + 3t - 10 = 0\)M1 Attempts to multiply by \(1-t^2\) and \(1+t^2\), expands and simplifies to a cubic in \(t\)
\(\Rightarrow (t+2)(\text{"6"}t^2 + \ldots t + \ldots) = 0\)M1 Attempts to take factor \((t+2)\) out of cubic. May be done by inspection or long division; must obtain correct coefficient for \(t^2\) and reach a 3-term quadratic
\(\Rightarrow (t+2)(6t^2+8t-10)=0\) or \((t+2)(3t^2+4t-5)=0\)A1 Correct equation. Must see equation written down not just values for \(a\), \(b\) and \(c\)
Part (c):
AnswerMarks Guidance
WorkingMark Guidance
\(t+2=0 \Rightarrow x = 2\times\arctan(-2) = \ldots\)M1 Attempts to solve \(t+2=0\), look for attempt at \(\arctan(\pm 2)\) and attempt to double. Allow in radians
\(x = -126.86\ldots°\) (allow awrt \(-127°\))A1 awrt \(-127°\)
\(3t^2+4t-5=0 \Rightarrow t = \frac{-4 \pm \sqrt{16-4\times3\times-5}}{6} = \frac{-2\pm\sqrt{19}}{3}\) \((= 0.786\ldots, -2.11\ldots)\) \(\Rightarrow x = 2\times\arctan(\ldots) = \ldots\)M1 Solves quadratic and attempts arctan and doubles. Usual rules apply for solving quadratic
\(x = \text{(awrt)} -129°, 76.4°\) \((-129.486\ldots, 76.355\ldots)\)A1 (one), A1 (both) Either awrt \(-129°\) or awrt \(76.4°\) for first A1; both awrt \(-129°\) and awrt \(76.4°\) for second A1. No other values in range. Values outside range can be ignored
# Question 4(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}$ or $2\cos^2\frac{x}{2}-1$ or $1-2\sin^2\frac{x}{2}$ | B1 | Any correct appropriate double angle identity for $\cos x$ or possibly $\tan x$ |
| $\Rightarrow \cos x = \left(\frac{1}{\sqrt{1+t^2}}\right)^2-\left(\frac{t}{\sqrt{1+t^2}}\right)^2$ | M1 | Substitutes correct expressions for $\cos\frac{x}{2}$ and/or $\sin\frac{x}{2}$ in terms of $t$ |
| $= \frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}$ * | A1* | Completes proof with sufficient working shown, no errors; given mark |

# Question 4 (Trigonometry with t-substitution):

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $= \frac{2t}{1+t^2} \times \frac{1-t^2}{2t} = \frac{1-t^2}{1+t^2}$ | M1 | Substituting correct expressions for $\cos\frac{x}{2}$, $\sin\frac{x}{2}$ and $\tan\frac{x}{2}$ in terms of $t$ |
| Completion to $\frac{1-t^2}{1+t^2}$ | A1* | Completing the proof with sufficient working shown, no errors. Allow $\theta$ for $x$ for first 2 marks but must obtain $\cos x = \ldots$ for A1* |

**Special Case** (quoting $\tan x$ and/or $\sin x$ to verify): Scores SC B1 only

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $3\tan x - 10\cos x = 10 \Rightarrow 3 \times \frac{2t}{1-t^2} - 10 \times \frac{1-t^2}{1+t^2} = 10$ | B1 | Applies part (a) and half angle formula for $\tan$ to give correct equation in $t$ |
| $\Rightarrow 6t(1+t^2) - 10(1-t^2)^2 = 10(1-t^2)(1+t^2)$ leading to $6t^3 + 20t^2 + 6t - 20 = 0$ or $3t^3 + 10t^2 + 3t - 10 = 0$ | M1 | Attempts to multiply by $1-t^2$ and $1+t^2$, expands and simplifies to a cubic in $t$ |
| $\Rightarrow (t+2)(\text{"6"}t^2 + \ldots t + \ldots) = 0$ | M1 | Attempts to take factor $(t+2)$ out of cubic. May be done by inspection or long division; must obtain correct coefficient for $t^2$ and reach a 3-term quadratic |
| $\Rightarrow (t+2)(6t^2+8t-10)=0$ or $(t+2)(3t^2+4t-5)=0$ | A1 | Correct equation. Must see equation written down not just values for $a$, $b$ and $c$ |

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| $t+2=0 \Rightarrow x = 2\times\arctan(-2) = \ldots$ | M1 | Attempts to solve $t+2=0$, look for attempt at $\arctan(\pm 2)$ and attempt to double. Allow in radians |
| $x = -126.86\ldots°$ (allow awrt $-127°$) | A1 | awrt $-127°$ |
| $3t^2+4t-5=0 \Rightarrow t = \frac{-4 \pm \sqrt{16-4\times3\times-5}}{6} = \frac{-2\pm\sqrt{19}}{3}$ $(= 0.786\ldots, -2.11\ldots)$ $\Rightarrow x = 2\times\arctan(\ldots) = \ldots$ | M1 | Solves quadratic and attempts arctan and doubles. Usual rules apply for solving quadratic |
| $x = \text{(awrt)} -129°, 76.4°$ $(-129.486\ldots, 76.355\ldots)$ | A1 (one), A1 (both) | Either awrt $-129°$ or awrt $76.4°$ for first A1; both awrt $-129°$ and awrt $76.4°$ for second A1. No other values in range. Values outside range can be ignored |

---
\begin{enumerate}
  \item (a) Given that $t = \tan \frac { X } { 2 }$ prove that
\end{enumerate}

$$\cos x \equiv \frac { 1 - t ^ { 2 } } { 1 + t ^ { 2 } }$$

(b) Show that the equation

$$3 \tan x - 10 \cos x = 10$$

can be written in the form

$$( t + 2 ) \left( a t ^ { 2 } + b t + c \right) = 0$$

where $t = \tan \frac { X } { 2 }$ and $a , b$ and $c$ are integers to be determined.\\
(c) Hence solve, for $- 180 ^ { \circ } < x < 180 ^ { \circ }$, the equation

$$3 \tan x - 10 \cos x = 10$$

\hfill \mbox{\textit{Edexcel FP1 AS 2024 Q4 [12]}}