OCR MEI C4 — Question 3 7 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeFind exact trigonometric values
DifficultyModerate -0.8 This question requires drawing standard special triangles (45-45-90 and 30-60-90) to find basic exact values, then applying the addition formula tan(A+B). While it involves multiple steps, these are all standard textbook techniques with no problem-solving insight required. The 'show that' structure provides guidance, making it easier than average but not trivial since it requires correct application of the tan addition formula.
Spec1.05g Exact trigonometric values: for standard angles1.05l Double angle formulae: and compound angle formulae
3 Using appropriate right-angled triangles, show that \(\tan 45 ^ { \circ } = 1\) and \(\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }\).
Hence show that \(\tan 75 ^ { \circ } = 2 + \sqrt { 3 }\).
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
Diagram: 45° triangle with sides 1, 1, \(\sqrt{2}\); \(\tan 45° = 1/1 = 1\)*B1 Need \(\sqrt{2}\) or indication that triangle is isosceles oe
Diagram: 30° triangle with sides 1, 2, \(\sqrt{3}\); \(\tan 30° = 1/\sqrt{3}\)*B1 Need all three sides oe. For both B marks AG so need to be convinced and need triangles but further explanation need not be on their diagram. Any given lengths must be consistent.
\(\tan 75° = \tan(45° + 30°)\)M1 Use of correct compound angle formula with \(45°, 30°\) soi
\(= \frac{\tan 45 + \tan 30}{1 - \tan 45\tan 30} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}}\)A1 Substitution in terms of \(\sqrt{3}\) in any correct form
\(= \frac{1 + \sqrt{3}}{-1 + \sqrt{3}}\)M1 Eliminating fractions within a fraction (or rationalising, whichever comes first) provided compound angle formula is used as \(\tan(A+B) = \tan(A\pm B)/(1\pm\tan A\tan B)\)
\(= \frac{(1+\sqrt{3})^2}{3-1}\) (oe eg \(\frac{3+\sqrt{3}}{3-\sqrt{3}} = \frac{(3+\sqrt{3})^2}{9-3}\))M1 Rationalising denominator (or eliminating fractions whichever comes second)
\(= \frac{(3 + 2\sqrt{3} + 1)}{3-1} = 2 + \sqrt{3}\)*A1 Correct only, AG so need to see working
[7] 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram: 45° triangle with sides 1, 1, $\sqrt{2}$; $\tan 45° = 1/1 = 1$* | B1 | Need $\sqrt{2}$ or indication that triangle is isosceles oe |
| Diagram: 30° triangle with sides 1, 2, $\sqrt{3}$; $\tan 30° = 1/\sqrt{3}$* | B1 | Need all three sides oe. For both B marks **AG** so need to be convinced and need triangles but further explanation need not be on their diagram. Any given lengths must be consistent. |
| $\tan 75° = \tan(45° + 30°)$ | M1 | Use of **correct** compound angle formula with $45°, 30°$ soi |
| $= \frac{\tan 45 + \tan 30}{1 - \tan 45\tan 30} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}}$ | A1 | Substitution in terms of $\sqrt{3}$ in any **correct** form |
| $= \frac{1 + \sqrt{3}}{-1 + \sqrt{3}}$ | M1 | Eliminating fractions within a fraction (or rationalising, whichever comes first) provided compound angle formula is used as $\tan(A+B) = \tan(A\pm B)/(1\pm\tan A\tan B)$ |
| $= \frac{(1+\sqrt{3})^2}{3-1}$ (oe eg $\frac{3+\sqrt{3}}{3-\sqrt{3}} = \frac{(3+\sqrt{3})^2}{9-3}$) | M1 | Rationalising denominator (or eliminating fractions whichever comes second) |
| $= \frac{(3 + 2\sqrt{3} + 1)}{3-1} = 2 + \sqrt{3}$* | A1 | **Correct** only, **AG** so need to see working |
| | **[7]** | |

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3 Using appropriate right-angled triangles, show that $\tan 45 ^ { \circ } = 1$ and $\tan 30 ^ { \circ } = \frac { 1 } { \sqrt { 3 } }$.\\
Hence show that $\tan 75 ^ { \circ } = 2 + \sqrt { 3 }$.

\hfill \mbox{\textit{OCR MEI C4  Q3 [7]}}
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