OCR MEI Paper 3 2023 June — Question 1 3 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeGiven one function find others
DifficultyEasy -1.2 This is a straightforward application of the Pythagorean identity sin²θ + cos²θ = 1, requiring only substitution and careful attention to the sign (obtuse angle means negative cosine). It's a standard textbook exercise with minimal problem-solving demand, easier than average A-level questions.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1
1 In this question you must show detailed reasoning.
The obtuse angle \(\theta\) is such that \(\sin \theta = \frac { 2 } { \sqrt { 13 } }\).
Find the exact value of \(\cos \theta\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{4}{13} + \cos^2\theta = 1\) oeM1 M0 if decimal angles (d, r, g) seen unless superceded by correct method
\(\cos^2\theta = \frac{9}{13}\)A1
\(\cos\theta = -\frac{3}{\sqrt{13}}\)A1 OR \(\cos\theta = -\frac{3\sqrt{13}}{13}\); If decimals seen (as check) isw
Alternative method: \(13 = 4 + a^2\) or correct triangle \((2, 3, \sqrt{13})\) seenM1
\(a = (\pm)3\) soiA1
\(\cos\theta = -\frac{3}{\sqrt{13}}\)A1
[3] 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{13} + \cos^2\theta = 1$ oe | M1 | **M0** if decimal angles (d, r, g) seen unless superceded by correct method |
| $\cos^2\theta = \frac{9}{13}$ | A1 | |
| $\cos\theta = -\frac{3}{\sqrt{13}}$ | A1 | OR $\cos\theta = -\frac{3\sqrt{13}}{13}$; If decimals seen (as check) isw |
| **Alternative method:** $13 = 4 + a^2$ or correct triangle $(2, 3, \sqrt{13})$ seen | M1 | |
| $a = (\pm)3$ soi | A1 | |
| $\cos\theta = -\frac{3}{\sqrt{13}}$ | A1 | |
| **[3]** | | |

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1 In this question you must show detailed reasoning.\\
The obtuse angle $\theta$ is such that $\sin \theta = \frac { 2 } { \sqrt { 13 } }$.\\
Find the exact value of $\cos \theta$.

\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q1 [3]}}
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