| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Expand compound angle then solve |
| Difficulty | Standard +0.3 This is a standard compound angle question requiring expansion of sin(x+π/6) and cos(x-π/4) using addition formulae, then algebraic manipulation to reach tan x form. Part (b) is routine solving once the form is established. Slightly above average due to the algebraic manipulation needed, but follows a well-practiced technique with no novel insight required. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sin x \cos\frac{\pi}{6} + \cos x \sin\frac{\pi}{6} = \cos x \cos\frac{\pi}{4} + \sin x \sin\frac{\pi}{4}\) | M1 | Using a compound angle formula at least once |
| \(\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}\cos x + \frac{\sqrt{2}}{2}\sin x\) | M1 | Uses exact values for at least 2 trigonometric terms |
| \(\sin x\left(\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\right) = \cos x\left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right)\) | M1 | Collecting terms and factorising |
| \(\tan x = \frac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\) | E1 | Complete argument with proper use of brackets where necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \frac{7\pi}{24}, \frac{31\pi}{24}\) | B1 | Both values without working and no others in range \(0 \leq x \leq 2\pi\). Allow decimal equivalents 0.916, 4.06 or better |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sin x \cos\frac{\pi}{6} + \cos x \sin\frac{\pi}{6} = \cos x \cos\frac{\pi}{4} + \sin x \sin\frac{\pi}{4}$ | M1 | Using a compound angle formula at least once |
| $\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}\cos x + \frac{\sqrt{2}}{2}\sin x$ | M1 | Uses exact values for at least 2 trigonometric terms |
| $\sin x\left(\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\right) = \cos x\left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right)$ | M1 | Collecting terms and factorising |
| $\tan x = \frac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}$ | E1 | Complete argument with proper use of brackets where necessary |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \frac{7\pi}{24}, \frac{31\pi}{24}$ | B1 | Both values without working and no others in range $0 \leq x \leq 2\pi$. Allow decimal equivalents 0.916, 4.06 or better |
---6
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)$ can be written in the form $\tan x = \frac { \sqrt { 2 } - 1 } { \sqrt { 3 } - \sqrt { 2 } }$.
\item Hence solve the equation $\sin \left( x + \frac { 1 } { 6 } \pi \right) = \cos \left( x - \frac { 1 } { 4 } \pi \right)$ for $0 \leqslant x \leqslant 2 \pi$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q6 [5]}}