| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation only (no integral) |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard double angle and addition formulae. Part (a) requires routine application of sin 2A formula with quadrant consideration. Part (b)(i) is a standard 'show that' identity using compound angle formulae. Part (b)(ii) combines the identity with basic differentiation. All techniques are textbook exercises with clear pathways and no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Method for finding \(\sin A\) | M1 | |
| \(\sin A = -\dfrac{\sqrt{7}}{4}\) | A1 A1 | First A1 for \(\dfrac{\sqrt{7}}{4}\), exact. Second A1 for sign (even if dec. answer given) |
| Use of \(\sin 2A \equiv 2\sin A \cos A\) | M1 | |
| \(\sin 2A = -\dfrac{3\sqrt{7}}{8}\) or equivalent exact | A1\(\checkmark\) | (5) \(\pm\) f.t. Requires exact value, dependent on 2nd M |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\!\left(2x+\dfrac{\pi}{3}\right)+\cos\!\left(2x-\dfrac{\pi}{3}\right) \equiv \cos 2x\cos\dfrac{\pi}{3}-\sin 2x\sin\dfrac{\pi}{3}+\cos 2x\cos\dfrac{\pi}{3}+\sin 2x\sin\dfrac{\pi}{3}\) | M1 | This can be just written down (using factor formulae) for M1A1 |
| \(\equiv 2\cos 2x\cos\dfrac{\pi}{3}\) | A1 | M1A1 earned, if \(\equiv 2\cos 2x\cos\dfrac{\pi}{3}\) just written down, using factor theorem |
| \(\equiv \cos 2x\) AG | A1* | (3) Final A1* requires some working after first result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\sin x\cos x - 2\sin 2x\) | B1 B1 | First B1 for \(6\sin x\cos x\); second B1 for remaining term(s) |
| or \(\quad 6\sin x\cos x - 2\sin\!\left(2x+\dfrac{\pi}{3}\right)-2\sin\!\left(2x-\dfrac{\pi}{3}\right)\) | ||
| \(= 3\sin 2x - 2\sin 2x\) | M1 | |
| \(= \sin 2x\) AG | A1* | (4) |
| (12 marks) |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Method for finding $\sin A$ | M1 | |
| $\sin A = -\dfrac{\sqrt{7}}{4}$ | A1 A1 | First A1 for $\dfrac{\sqrt{7}}{4}$, exact. Second A1 for sign (even if dec. answer given) |
| Use of $\sin 2A \equiv 2\sin A \cos A$ | M1 | |
| $\sin 2A = -\dfrac{3\sqrt{7}}{8}$ or equivalent exact | A1$\checkmark$ | (5) $\pm$ f.t. Requires exact value, dependent on 2nd M |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\!\left(2x+\dfrac{\pi}{3}\right)+\cos\!\left(2x-\dfrac{\pi}{3}\right) \equiv \cos 2x\cos\dfrac{\pi}{3}-\sin 2x\sin\dfrac{\pi}{3}+\cos 2x\cos\dfrac{\pi}{3}+\sin 2x\sin\dfrac{\pi}{3}$ | M1 | This can be just written down (using factor formulae) for M1A1 |
| $\equiv 2\cos 2x\cos\dfrac{\pi}{3}$ | A1 | M1A1 earned, if $\equiv 2\cos 2x\cos\dfrac{\pi}{3}$ just written down, using factor theorem |
| $\equiv \cos 2x$ AG | A1* | (3) Final A1* requires some working after first result |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\sin x\cos x - 2\sin 2x$ | B1 B1 | First B1 for $6\sin x\cos x$; second B1 for remaining term(s) |
| or $\quad 6\sin x\cos x - 2\sin\!\left(2x+\dfrac{\pi}{3}\right)-2\sin\!\left(2x-\dfrac{\pi}{3}\right)$ | | |
| $= 3\sin 2x - 2\sin 2x$ | M1 | |
| $= \sin 2x$ AG | A1* | (4) |
| | | **(12 marks)** |\begin{enumerate}
\item (a) Given that $\cos A = \frac { 3 } { 4 }$, where $270 ^ { \circ } < A < 360 ^ { \circ }$, find the exact value of $\sin 2 A$.\\
(b) (i) Show that $\cos \left( 2 x + \frac { \pi } { 3 } \right) + \cos \left( 2 x - \frac { \pi } { 3 } \right) \equiv \cos 2 x$.
\end{enumerate}
Given that
$$y = 3 \sin ^ { 2 } x + \cos \left( 2 x + \frac { \pi } { 3 } \right) + \cos \left( 2 x - \frac { \pi } { 3 } \right)$$
(ii) show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x$.\\
\hfill \mbox{\textit{Edexcel C3 2006 Q8 [12]}}