OCR MEI C4 — Question 3 6 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeExpand compound angle then solve
DifficultyStandard +0.3 This is a straightforward compound angle formula question requiring expansion of sin(θ+45°), algebraic manipulation to reach the given form, then solving a linear trigonometric equation. The steps are standard and methodical with no novel insight required, making it slightly easier than average for A-level.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\sin(x+45°) = \sin x\cos 45° + \cos x\sin 45°\)M1 Use of correct compound angle formula
\(= \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\sin x + \cos x) = 2\cos x\)A1
\(\Rightarrow \sin x + \cos x = 2\sqrt{2}\cos x\)A1 Since AG, \(\sin x\cos 45° + \cos x\sin 45° = 2\cos x\); \(\sin x + \cos x = 2\sqrt{2}\cos x\) only gets M1. Need second line or statement of \(\cos 45° = \sin 45° = 1/\sqrt{2}\) oe as intermediate step to get A1 A1.
\(\Rightarrow \sin x = (2\sqrt{2}-1)\cos x\)M1 Terms collected and \(\tan x = \sin x/\cos x\) used
\(\Rightarrow \tan x = 2\sqrt{2}-1\)
\(\Rightarrow x = 61.32°\)A1 First correct solution
\(\quad\quad\quad 241.32°\)A1 Second correct solution and no others in range. 2dp but allow overspecification. Ignore solutions outside range. SC A1 for both 61.3° and 241.3°; SC A1 for both 1.07 and 4.21 radians (or better); SC A1 for incorrect answers rounding to 61.3° and 180°+ their ans. Do not award SC if extra solutions in range.
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin(x+45°) = \sin x\cos 45° + \cos x\sin 45°$ | M1 | Use of **correct** compound angle formula |
| $= \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\sin x + \cos x) = 2\cos x$ | A1 | |
| $\Rightarrow \sin x + \cos x = 2\sqrt{2}\cos x$ | A1 | Since AG, $\sin x\cos 45° + \cos x\sin 45° = 2\cos x$; $\sin x + \cos x = 2\sqrt{2}\cos x$ only gets M1. Need second line or statement of $\cos 45° = \sin 45° = 1/\sqrt{2}$ oe as intermediate step to get A1 A1. |
| $\Rightarrow \sin x = (2\sqrt{2}-1)\cos x$ | M1 | Terms collected and $\tan x = \sin x/\cos x$ used |
| $\Rightarrow \tan x = 2\sqrt{2}-1$ | | |
| $\Rightarrow x = 61.32°$ | A1 | First correct solution |
| $\quad\quad\quad 241.32°$ | A1 | Second correct solution and no others in range. 2dp but allow overspecification. Ignore solutions outside range. SC A1 for both 61.3° and 241.3°; SC A1 for both 1.07 and 4.21 radians (or better); SC A1 for incorrect answers rounding to 61.3° and 180°+ their ans. Do not award SC if extra solutions in range. |

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3 Given the equation $\sin \left( + 45 ^ { \circ } \right) = 2 \cos [$, show that $\sin + \cos = 22 \cos$. Hence solve, correct to 2 decimal places, the equation for $0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.$.

$$\leqslant \leqslant$$

\hfill \mbox{\textit{OCR MEI C4  Q3 [6]}}