OCR MEI AS Paper 2 2021 November — Question 3 3 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2021
SessionNovember
Marks3
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Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypePeriodicity and symmetry of trig functions
DifficultyModerate -0.8 This question requires understanding periodicity of tan (period 180°) and recognizing that 690° = 720° - 30° = 4(180°) - 30°, making tan 690° = tan(-30°) = -tan 30°. While it requires explanation rather than just calculation, the concepts are fundamental and the arithmetic is straightforward, making it easier than average for AS-level.
Spec1.05a Sine, cosine, tangent: definitions for all arguments1.05f Trigonometric function graphs: symmetries and periodicities
3 In this question you must show detailed reasoning. You are given that \(\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }\).
Explain why \(\tan 690 ^ { \circ } = - \frac { 1 } { \sqrt { 3 } }\).
Question 3:
AnswerMarks Guidance
\(\tan(-30°) = -\frac{1}{\sqrt{3}}\)B1 Marks may be gained in a different order; allow for any correct method.
\(\tan(-30° + 4 \times 180°) = \tan(690°)\)B1 Or \(\tan(-30° + 2 \times 360°) = \tan 690°\); Accept \(-30° + 4 \times 180° = 690°\)
completion to \(\tan(690°) = -\frac{1}{\sqrt{3}}\)B1
[3]
**Question 3:**

$\tan(-30°) = -\frac{1}{\sqrt{3}}$ | B1 | Marks may be gained in a different order; allow for any correct method.

$\tan(-30° + 4 \times 180°) = \tan(690°)$ | B1 | Or $\tan(-30° + 2 \times 360°) = \tan 690°$; Accept $-30° + 4 \times 180° = 690°$

completion to $\tan(690°) = -\frac{1}{\sqrt{3}}$ | B1 |

**[3]**

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3 In this question you must show detailed reasoning.
You are given that $\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }$.\\
Explain why $\tan 690 ^ { \circ } = - \frac { 1 } { \sqrt { 3 } }$.

\hfill \mbox{\textit{OCR MEI AS Paper 2 2021 Q3 [3]}}
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Q1 3 Q2 2 Q3 3 Q4 4 Q5 7 Q6 6 Q7 7 Q8 4 Q9 5 Q10 6 Q11 6 Q12 8 Q13 9