OCR C3 2008 January — Question 3 7 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2008
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeMultiple angle equations
DifficultyModerate -0.3 This question tests routine manipulation of reciprocal trig functions and standard solving techniques. Part (a) requires converting sec to cos and handling the half-angle (straightforward substitution), while part (b) uses the identity tan·cot=1 to reach tan²β=7. Both are standard textbook exercises with clear methods and no novel problem-solving required, making them slightly easier than average A-level questions.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

3
  1. Solve, for \(0 ^ { \circ } < \alpha < 180 ^ { \circ }\), the equation \(\sec \frac { 1 } { 2 } \alpha = 4\).
  2. Solve, for \(0 ^ { \circ } < \beta < 180 ^ { \circ }\), the equation \(\tan \beta = 7 \cot \beta\).

3
\begin{enumerate}[label=(\alph*)]
\item Solve, for $0 ^ { \circ } < \alpha < 180 ^ { \circ }$, the equation $\sec \frac { 1 } { 2 } \alpha = 4$.
\item Solve, for $0 ^ { \circ } < \beta < 180 ^ { \circ }$, the equation $\tan \beta = 7 \cot \beta$.
\end{enumerate}

\hfill \mbox{\textit{OCR C3 2008 Q3 [7]}}