Question 6 - A-Level Maths

Edexcel C3 2006 June — Question 6 10 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2006
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Solve equation with reciprocal functions
Difficulty	Standard +0.3 This is a structured multi-part question on reciprocal trig identities. Part (a) is routine derivation from a standard identity, part (b) uses difference of squares (guided by 'hence'), and part (c) applies the result to solve an equation in a restricted domain. While it requires multiple steps, each part is scaffolded and uses standard C3 techniques without requiring novel insight.
Spec	1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 1.05o Trigonometric equations: solve in given intervals 1.05p Proof involving trig: functions and identities

(a) Using $\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1$, show that $\operatorname { cosec } ^ { 2 } \theta - \cot ^ { 2 } \theta \equiv 1$.
(b) Hence, or otherwise, prove that

$$\operatorname { cosec } ^ { 4 } \theta - \cot ^ { 4 } \theta \equiv \operatorname { cosec } ^ { 2 } \theta + \cot ^ { 2 } \theta$$ (c) Solve, for $90 ^ { \circ } < \theta < 180 ^ { \circ }$, $$\operatorname { cosec } ^ { 4 } \theta - \cot ^ { 4 } \theta = 2 - \cot \theta$$

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Question 6:

Part (a)

Answer	Marks
Dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\sin^2\theta$: $\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}$	M1
Completion: $1 + \cot^2\theta \equiv \cosec^2\theta \Rightarrow \cosec^2\theta - \cot^2\theta \equiv 1$	A1* (2)

Part (b)

Answer	Marks
$\cosec^4\theta - \cot^4\theta \equiv (\cosec^2\theta - \cot^2\theta)(\cosec^2\theta + \cot^2\theta)$	M1
$\equiv (\cosec^2\theta + \cot^2\theta)$ using (a)	A1* (2)

Part (c)

Answer	Marks	Guidance
Using (b) to form: $\cosec^2\theta + \cot^2\theta \equiv 2 - \cot\theta$	M1
Forming quadratic in $\cot\theta$: $1 + \cot^2\theta + \cot^2\theta \equiv 2 - \cot\theta$ {using (a)}	M1
$2\cot^2\theta + \cot\theta - 1 = 0$	A1
Solving: $(2\cot\theta - 1)(\cot\theta + 1) = 0$	M1
$\cot\theta = \frac{1}{2}$ or $\cot\theta = -1$	A1
$\theta = 135°$ (or correct value(s) for candidate dep. on 3Ms)	A1$\sqrt{}$ (6)	Ignore solutions outside range. Extra solutions in range loses A1$\sqrt{}$

# Question 6:

## Part (a)
Dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\sin^2\theta$: $\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}$ | M1 |

Completion: $1 + \cot^2\theta \equiv \cosec^2\theta \Rightarrow \cosec^2\theta - \cot^2\theta \equiv 1$ | A1* (2) |

## Part (b)
$\cosec^4\theta - \cot^4\theta \equiv (\cosec^2\theta - \cot^2\theta)(\cosec^2\theta + \cot^2\theta)$ | M1 |

$\equiv (\cosec^2\theta + \cot^2\theta)$ using (a) | A1* (2) |

## Part (c)
Using (b) to form: $\cosec^2\theta + \cot^2\theta \equiv 2 - \cot\theta$ | M1 |

Forming quadratic in $\cot\theta$: $1 + \cot^2\theta + \cot^2\theta \equiv 2 - \cot\theta$ {using (a)} | M1 |

$2\cot^2\theta + \cot\theta - 1 = 0$ | A1 |

Solving: $(2\cot\theta - 1)(\cot\theta + 1) = 0$ | M1 |

$\cot\theta = \frac{1}{2}$ or $\cot\theta = -1$ | A1 |

$\theta = 135°$ (or correct value(s) for candidate dep. on 3Ms) | A1$\sqrt{}$ (6) | Ignore solutions outside range. Extra solutions in range loses A1$\sqrt{}$

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Show LaTeX source

\begin{enumerate}
  \item (a) Using $\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1$, show that $\operatorname { cosec } ^ { 2 } \theta - \cot ^ { 2 } \theta \equiv 1$.\\
(b) Hence, or otherwise, prove that
\end{enumerate}

$$\operatorname { cosec } ^ { 4 } \theta - \cot ^ { 4 } \theta \equiv \operatorname { cosec } ^ { 2 } \theta + \cot ^ { 2 } \theta$$

(c) Solve, for $90 ^ { \circ } < \theta < 180 ^ { \circ }$,

$$\operatorname { cosec } ^ { 4 } \theta - \cot ^ { 4 } \theta = 2 - \cot \theta$$

\hfill \mbox{\textit{Edexcel C3 2006 Q6 [10]}}

This paper (8 questions)

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Q1 6 Q2 6 Q3 9 Q4 9 Q5 11 Q6 10 Q7 12 Q8 12

Answer	Marks
Dividing \(\sin^2\theta + \cos^2\theta \equiv 1\) by \(\sin^2\theta\): \(\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta}\)	M1
Completion: \(1 + \cot^2\theta \equiv \cosec^2\theta \Rightarrow \cosec^2\theta - \cot^2\theta \equiv 1\)	A1* (2)

Answer	Marks
\(\cosec^4\theta - \cot^4\theta \equiv (\cosec^2\theta - \cot^2\theta)(\cosec^2\theta + \cot^2\theta)\)	M1
\(\equiv (\cosec^2\theta + \cot^2\theta)\) using (a)	A1* (2)

Answer	Marks	Guidance
Using (b) to form: \(\cosec^2\theta + \cot^2\theta \equiv 2 - \cot\theta\)	M1
Forming quadratic in \(\cot\theta\): \(1 + \cot^2\theta + \cot^2\theta \equiv 2 - \cot\theta\) {using (a)}	M1
\(2\cot^2\theta + \cot\theta - 1 = 0\)	A1
Solving: \((2\cot\theta - 1)(\cot\theta + 1) = 0\)	M1
\(\cot\theta = \frac{1}{2}\) or \(\cot\theta = -1\)	A1
\(\theta = 135°\) (or correct value(s) for candidate dep. on 3Ms)	A1\(\sqrt{}\) (6)	Ignore solutions outside range. Extra solutions in range loses A1\(\sqrt{}\)