OCR MEI C4 2008 January — Question 6 3 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2008
SessionJanuary
Marks3
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TopicReciprocal Trig & Identities
TypeSolve equation with reciprocal functions
DifficultyEasy -1.2 This is a straightforward recall question requiring only the definition of cosec (1/sin) and basic inverse sine calculation. Students simply convert to sin θ = 1/3, find the reference angle, and identify two solutions in the given range using CAST diagram—a routine exercise with no problem-solving element.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05o Trigonometric equations: solve in given intervals
6 Solve the equation \(\operatorname { cosec } \theta = 3\), for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
AnswerMarks
\(\frac{\phi + 1}{2\phi - 1} = \frac{\frac{1 + \sqrt{5}}{2} + 1}{1 + \sqrt{5} - 1}\)M1 A1
\(= \frac{\frac{3 + \sqrt{5}}{2}}{2\sqrt{5}}\)
AnswerMarks
\(= \frac{(3 + \sqrt{5})\sqrt{5}}{2\sqrt{5} \times \sqrt{5}} = \frac{3\sqrt{5} + 5}{10}\)E1 
$\frac{\phi + 1}{2\phi - 1} = \frac{\frac{1 + \sqrt{5}}{2} + 1}{1 + \sqrt{5} - 1}$ | M1 A1 |

$= \frac{\frac{3 + \sqrt{5}}{2}}{2\sqrt{5}}$

$= \frac{(3 + \sqrt{5})\sqrt{5}}{2\sqrt{5} \times \sqrt{5}} = \frac{3\sqrt{5} + 5}{10}$ | E1 |
6 Solve the equation $\operatorname { cosec } \theta = 3$, for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.

\hfill \mbox{\textit{OCR MEI C4 2008 Q6 [3]}}
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Q1 7 Q2 8 Q3 5 Q4 7 Q5 6 Q6 3 Q7 18 Q8 18