WJEC Unit 3 2018 June — Question 13 8 marks

Exam BoardWJEC
ModuleUnit 3 (Unit 3)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeRange of rational function with harmonic denominator
DifficultyStandard +0.3 This is a standard R-formula question with three routine parts: (a) converting to R sin(θ-α) form using Pythagorean theorem and tan α, (b) solving a trigonometric equation using the result from (a), and (c) finding max/min of a reciprocal function. All parts follow well-established procedures taught in C3/C4 with no novel problem-solving required, making it slightly easier than average.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

  1. Express \(8\sin\theta - 15\cos\theta\) in the form \(R\sin(\theta - \alpha)\), where \(R\) and \(\alpha\) are constants with \(R > 0\) and \(0° < \alpha < 90°\). [3]
  2. Find all values of \(\theta\) in the range \(0° < \theta < 360°\) satisfying $$8\sin\theta - 15\cos\theta - 7 = 0.$$ [3]
  3. Determine the greatest value and the least value of the expression $$\frac{1}{8\sin\theta - 15\cos\theta + 23}.$$ [2]

\begin{enumerate}[label=(\alph*)]
\item Express $8\sin\theta - 15\cos\theta$ in the form $R\sin(\theta - \alpha)$, where $R$ and $\alpha$ are constants with $R > 0$ and $0° < \alpha < 90°$. [3]

\item Find all values of $\theta$ in the range $0° < \theta < 360°$ satisfying
$$8\sin\theta - 15\cos\theta - 7 = 0.$$ [3]

\item Determine the greatest value and the least value of the expression
$$\frac{1}{8\sin\theta - 15\cos\theta + 23}.$$ [2]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 3 2018 Q13 [8]}}