| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | First principles: trigonometric functions |
| Difficulty | Standard +0.8 This is a standard first principles differentiation proof for a trigonometric function, requiring multiple steps including the difference quotient, compound angle formula, algebraic manipulation, and application of two standard limits. While it's a bookwork proof that strong students memorize, it requires careful execution and understanding of limit properties, making it moderately challenging but still within the scope of routine A-level further maths content. |
| Spec | 1.07h Differentiation from first principles: for sin(x) and cos(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct fraction \(\dfrac{\cos(\theta+h)-\cos\theta}{h}\) stated | B1 | 2.1 — Allow \((\theta+h)-\theta\) in denominator |
| Uses compound angle formula: \(\cos(\theta+h)=\cos\theta\cos h \pm \sin\theta\sin h\) | M1 | 1.1b |
| Achieves \(\dfrac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\) | A1 | 1.1b |
| \(=-\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta\) | ||
| As \(h\to 0\): \(-\dfrac{\sin h}{h}\sin\theta+\left(\dfrac{\cos h-1}{h}\right)\cos\theta \to -1\sin\theta+0\cos\theta\) | dM1 | 2.1 — Dependent on B and M; must isolate \(\dfrac{\sin h}{h}\) and \(\dfrac{\cos h-1}{h}\) and replace with 1 and 0 respectively |
| \(\therefore \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta)=-\sin\theta\) | A1* | 2.5 — Must use correct limiting argument language; do not award for simply setting \(h=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{d\theta}(\cos x) = \lim_{h \to 0}\left(-\frac{\sin h}{h}\sin\theta + \left(\frac{\cos h - 1}{h}\right)\cos\theta\right) = -1\sin\theta + 0\cos\theta = -\sin\theta\) | A1 | Final A1 for correct limiting argument in \(x\), followed by \(\frac{d}{d\theta}(\cos\theta) = -\sin\theta\) |
| Applying \(h \to 0\), \(\sin h \to h\), \(\cos h \to 1\): \(\lim_{h\to 0}\left(\frac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\right) = \frac{-\sin\theta(h)}{h} = -\sin\theta\) | M0A0 | Final M0A0 for incorrect application of limits |
| Alternative: \(\frac{\cos(\theta+h)-\cos(\theta-h)}{(\theta+h)-(\theta-h)}\) simplifies to \(\frac{-2\sin\theta\sin h}{2h}\) | — | Alternative method note |
# Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct fraction $\dfrac{\cos(\theta+h)-\cos\theta}{h}$ stated | B1 | 2.1 — Allow $(\theta+h)-\theta$ in denominator |
| Uses compound angle formula: $\cos(\theta+h)=\cos\theta\cos h \pm \sin\theta\sin h$ | M1 | 1.1b |
| Achieves $\dfrac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}$ | A1 | 1.1b |
| $=-\dfrac{\sin h}{h}\sin\theta + \left(\dfrac{\cos h - 1}{h}\right)\cos\theta$ | | |
| As $h\to 0$: $-\dfrac{\sin h}{h}\sin\theta+\left(\dfrac{\cos h-1}{h}\right)\cos\theta \to -1\sin\theta+0\cos\theta$ | dM1 | 2.1 — Dependent on B and M; must isolate $\dfrac{\sin h}{h}$ and $\dfrac{\cos h-1}{h}$ and replace with 1 and 0 respectively |
| $\therefore \dfrac{\mathrm{d}}{\mathrm{d}\theta}(\cos\theta)=-\sin\theta$ | A1* | 2.5 — Must use correct limiting argument language; do not award for simply setting $h=0$ |
# Question 9 (Continued Notes):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{d\theta}(\cos x) = \lim_{h \to 0}\left(-\frac{\sin h}{h}\sin\theta + \left(\frac{\cos h - 1}{h}\right)\cos\theta\right) = -1\sin\theta + 0\cos\theta = -\sin\theta$ | A1 | Final A1 for correct limiting argument in $x$, followed by $\frac{d}{d\theta}(\cos\theta) = -\sin\theta$ |
| Applying $h \to 0$, $\sin h \to h$, $\cos h \to 1$: $\lim_{h\to 0}\left(\frac{\cos\theta\cos h - \sin\theta\sin h - \cos\theta}{h}\right) = \frac{-\sin\theta(h)}{h} = -\sin\theta$ | M0A0 | Final M0A0 for incorrect application of limits |
| Alternative: $\frac{\cos(\theta+h)-\cos(\theta-h)}{(\theta+h)-(\theta-h)}$ simplifies to $\frac{-2\sin\theta\sin h}{2h}$ | — | Alternative method note |
---\begin{enumerate}
\item Given that $\theta$ is measured in radians, prove, from first principles, that
\end{enumerate}
$$\frac { \mathrm { d } } { \mathrm {~d} \theta } ( \cos \theta ) = - \sin \theta$$
You may assume the formula for $\cos ( A \pm B )$ and that as $h \rightarrow 0 , \frac { \sin h } { h } \rightarrow 1$ and $\frac { \cos h - 1 } { h } \rightarrow 0$\\
(5)
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q9 [5]}}