| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| 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 routine steps: converting to R sin(x+α) using standard formulas, reading off minimum value and location, then solving a transformed equation. All techniques are textbook exercises requiring no novel insight, though part (iii) requires careful handling of the double angle and multiple solutions within the given range. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(4\sin x + 3\cos x = R\sin x\cos\alpha + R\cos x\sin\alpha\) | ||
| \(R\cos\alpha = 4\), \(R\sin\alpha = 3\) | M1 | |
| \(\therefore R = \sqrt{4^2 + 3^2} = 5\) | A1 | |
| \(\tan\alpha = \frac{3}{4}\), \(\alpha = 0.644\) (3sf) | A1 | |
| \(\therefore 4\sin x + 3\cos x = 5\sin(x + 0.644)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| minimum \(= -5\) | B1 | |
| occurs when \(x + 0.6435 = \frac{3\pi}{2}\), \(x = 4.07\) (3sf) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(5\sin(2\theta + 0.6435) = 2\) | ||
| \(\sin(2\theta + 0.6435) = 0.4\) | M1 | |
| \(2\theta + 0.6435 = \pi - 0.4115\), \(2\pi + 0.4115\) | ||
| \(2\theta = 2.087, 6.051\) | M1 | |
| \(\theta = 1.04, 3.03\) (2dp) | A2 | (10) |
# Question 6:
## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $4\sin x + 3\cos x = R\sin x\cos\alpha + R\cos x\sin\alpha$ | | |
| $R\cos\alpha = 4$, $R\sin\alpha = 3$ | M1 | |
| $\therefore R = \sqrt{4^2 + 3^2} = 5$ | A1 | |
| $\tan\alpha = \frac{3}{4}$, $\alpha = 0.644$ (3sf) | A1 | |
| $\therefore 4\sin x + 3\cos x = 5\sin(x + 0.644)$ | | |
## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| minimum $= -5$ | B1 | |
| occurs when $x + 0.6435 = \frac{3\pi}{2}$, $x = 4.07$ (3sf) | M1 A1 | |
## Part (iii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $5\sin(2\theta + 0.6435) = 2$ | | |
| $\sin(2\theta + 0.6435) = 0.4$ | M1 | |
| $2\theta + 0.6435 = \pi - 0.4115$, $2\pi + 0.4115$ | | |
| $2\theta = 2.087, 6.051$ | M1 | |
| $\theta = 1.04, 3.03$ (2dp) | A2 | **(10)** |
---\begin{enumerate}
\item (i) Express $4 \sin x + 3 \cos x$ in the form $R \sin ( x + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.\\
(ii) State the minimum value of $4 \sin x + 3 \cos x$ and the smallest positive value of $x$ for which this minimum value occurs.\\
(iii) Solve the equation
\end{enumerate}
$$4 \sin 2 \theta + 3 \cos 2 \theta = 2$$
for $\theta$ in the interval $0 \leq \theta \leq \pi$, giving your answers to 2 decimal places.\\
\hfill \mbox{\textit{OCR C3 Q6 [10]}}