OCR MEI C4 2009 January — Question 4 3 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2009
SessionJanuary
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve algebraic trigonometric identity
DifficultyStandard +0.3 This is a straightforward algebraic proof requiring conversion of cotangent to cosine/sine ratios, finding a common denominator, and applying the sine difference formula. While it involves multiple steps, each is standard technique with no novel insight required, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05p Proof involving trig: functions and identities
4 Prove that \(\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(V = \frac{1}{2}\times80\times(40+5)\)M1
\(\times30\text{ cm}^3 = 54000\text{ cm}^3 = 54\) litresM1, A1 \(\times30\) 
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \frac{1}{2}\times80\times(40+5)$ | M1 | |
| $\times30\text{ cm}^3 = 54000\text{ cm}^3 = 54$ litres | M1, A1 | $\times30$ |
4 Prove that $\cot \beta - \cot \alpha = \frac { \sin ( \alpha - \beta ) } { \sin \alpha \sin \beta }$.

\hfill \mbox{\textit{OCR MEI C4 2009 Q4 [3]}}
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