| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question with scaffolded parts: (a) is routine identity proof, (b) is algebraic manipulation using part (a), (c) is textbook harmonic form conversion (R cos(θ-α)), and (d) applies the result to solve. All techniques are standard with clear signposting, making it slightly easier than average despite multiple parts. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A=B \Rightarrow \cos(A+A) = \cos 2A = \cos A\cos A - \sin A\sin A\) | M1 | Applies \(A=B\) to \(\cos(A+B)\) to give the underlined equation or \(\cos 2A = \cos^2 A - \sin^2 A\) |
| \(\cos 2A = \cos^2 A - \sin^2 A\) and \(\cos^2 A + \sin^2 A = 1\) gives \(\cos 2A = 1 - \sin^2 A - \sin^2 A = 1 - 2\sin^2 A\) | A1 AG | Complete proof, with a link between LHS and RHS. No errors seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C_1 = C_2 \Rightarrow 3\sin 2x = 4\sin^2 x - 2\cos 2x\) | M1 | Eliminating \(y\) correctly |
| \(3\sin 2x = 4\left(\frac{1-\cos 2x}{2}\right) - 2\cos 2x\) | M1 | Using result in part (a) to substitute for \(\sin^2 x\) as \(\frac{\pm 1 \pm \cos 2x}{2}\) or \(k\sin^2 x\) as \(k\left(\frac{\pm 1 \pm \cos 2x}{2}\right)\) to produce equation in only double angles |
| \(3\sin 2x = 2(1-\cos 2x) - 2\cos 2x\) | ||
| \(3\sin 2x + 4\cos 2x = 2\) | A1 AG | Rearranges to give correct result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\) | B1 | \(R = 5\) |
| \(\tan\alpha = \frac{3}{4} \Rightarrow \alpha = 36.86989...°\) | M1 | \(\tan\alpha = \pm\frac{3}{4}\) or \(\tan\alpha = \pm\frac{4}{3}\) or \(\sin\alpha = \pm\frac{3}{\text{their }R}\) or \(\cos\alpha = \pm\frac{4}{\text{their }R}\) |
| A1 | awrt 36.87 | |
| Hence \(3\sin 2x + 4\cos 2x = 5\cos(2x - 36.87)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos(2x - 36.87) = \frac{2}{5}\) | M1 | \(\cos(2x \pm \text{their }\alpha) = \frac{2}{\text{their }R}\) |
| \((2x - 36.87) = 66.42182...°\) | A1 | awrt 66 |
| \((2x - 36.87) = 360 - 66.42182...°\) | ||
| \(x = 51.64591...°,\ 165.22409...°\) | A1 | One of either awrt 51.6 or awrt 51.7 or awrt 165.2 or awrt 165.3 |
| A1 | Both awrt 51.6 AND awrt 165.2. If EXTRA solutions inside \(0 \leq x < 180°\) withhold final accuracy mark. Ignore EXTRA solutions outside \(0 \leq x < 180°\) |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=B \Rightarrow \cos(A+A) = \cos 2A = \cos A\cos A - \sin A\sin A$ | M1 | Applies $A=B$ to $\cos(A+B)$ to give the underlined equation or $\cos 2A = \cos^2 A - \sin^2 A$ |
| $\cos 2A = \cos^2 A - \sin^2 A$ and $\cos^2 A + \sin^2 A = 1$ gives $\cos 2A = 1 - \sin^2 A - \sin^2 A = 1 - 2\sin^2 A$ | A1 AG | Complete proof, with a link between LHS and RHS. No errors seen. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C_1 = C_2 \Rightarrow 3\sin 2x = 4\sin^2 x - 2\cos 2x$ | M1 | Eliminating $y$ correctly |
| $3\sin 2x = 4\left(\frac{1-\cos 2x}{2}\right) - 2\cos 2x$ | M1 | Using result in part (a) to substitute for $\sin^2 x$ as $\frac{\pm 1 \pm \cos 2x}{2}$ or $k\sin^2 x$ as $k\left(\frac{\pm 1 \pm \cos 2x}{2}\right)$ to produce equation in only double angles |
| $3\sin 2x = 2(1-\cos 2x) - 2\cos 2x$ | | |
| $3\sin 2x + 4\cos 2x = 2$ | A1 AG | Rearranges to give correct result |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ | B1 | $R = 5$ |
| $\tan\alpha = \frac{3}{4} \Rightarrow \alpha = 36.86989...°$ | M1 | $\tan\alpha = \pm\frac{3}{4}$ or $\tan\alpha = \pm\frac{4}{3}$ or $\sin\alpha = \pm\frac{3}{\text{their }R}$ or $\cos\alpha = \pm\frac{4}{\text{their }R}$ |
| | A1 | awrt 36.87 |
| Hence $3\sin 2x + 4\cos 2x = 5\cos(2x - 36.87)$ | | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos(2x - 36.87) = \frac{2}{5}$ | M1 | $\cos(2x \pm \text{their }\alpha) = \frac{2}{\text{their }R}$ |
| $(2x - 36.87) = 66.42182...°$ | A1 | awrt 66 |
| $(2x - 36.87) = 360 - 66.42182...°$ | | |
| $x = 51.64591...°,\ 165.22409...°$ | A1 | One of either awrt 51.6 or awrt 51.7 or awrt 165.2 or awrt 165.3 |
| | A1 | Both awrt 51.6 AND awrt 165.2. If EXTRA solutions inside $0 \leq x < 180°$ withhold final accuracy mark. Ignore EXTRA solutions outside $0 \leq x < 180°$ |
---\begin{enumerate}
\item (a) Use the identity $\cos ( A + B ) = \cos A \cos B - \sin A \sin B$, to show that
\end{enumerate}
$$\cos 2 A = 1 - 2 \sin ^ { 2 } A$$
The curves $C _ { 1 }$ and $C _ { 2 }$ have equations
$$\begin{aligned}
& C _ { 1 } : \quad y = 3 \sin 2 x \\
& C _ { 2 } : \quad y = 4 \sin ^ { 2 } x - 2 \cos 2 x
\end{aligned}$$
(b) Show that the $x$-coordinates of the points where $C _ { 1 }$ and $C _ { 2 }$ intersect satisfy the equation
$$4 \cos 2 x + 3 \sin 2 x = 2$$
(c) Express $4 \cos 2 x + 3 \sin 2 x$ in the form $R \cos ( 2 x - \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$, giving the value of $\alpha$ to 2 decimal places.\\
(d) Hence find, for $0 \leqslant x < 180 ^ { \circ }$, all the solutions of
$$4 \cos 2 x + 3 \sin 2 x = 2$$
giving your answers to 1 decimal place.\\
\hfill \mbox{\textit{Edexcel C3 2009 Q6 [12]}}