AQA C3 2011 June — Question 4 12 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
TopicReciprocal Trig & Identities
TypeMultiple angle equations
DifficultyStandard +0.3 This is a standard C3 trigonometry question testing cosecant and cotangent equations. Part (a)(i) is routine recall (2 marks), part (a)(ii) requires substitution using the identity cot²θ + 1 = cosec²θ to form a quadratic in cosec, then solving with domain considerations (6 marks), and part (b) tests understanding of function transformations. While multi-step, these are textbook techniques with no novel insight required, making it slightly easier than average overall.
Spec1.02w Graph transformations: simple transformations of f(x)1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05o Trigonometric equations: solve in given intervals

    1. Solve the equation \(\cosec \theta = -4\) for \(0° < \theta < 360°\), giving your answers to the nearest 0.1°. [2]
    2. Solve the equation $$2\cot^2(2x + 30°) = 2 - 7\cosec(2x + 30°)$$ for \(0° < x < 180°\), giving your answers to the nearest 0.1°. [6]
  1. Describe a sequence of two geometrical transformations that maps the graph of \(y = \cosec x\) onto the graph of \(y = \cosec(2x + 30°)\). [4]

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Solve the equation $\cosec \theta = -4$ for $0° < \theta < 360°$, giving your answers to the nearest 0.1°. [2]

\item Solve the equation
$$2\cot^2(2x + 30°) = 2 - 7\cosec(2x + 30°)$$
for $0° < x < 180°$, giving your answers to the nearest 0.1°. [6]
\end{enumerate}

\item Describe a sequence of two geometrical transformations that maps the graph of $y = \cosec x$ onto the graph of $y = \cosec(2x + 30°)$. [4]
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2011 Q4 [12]}}