Given that θ is measured in radians, prove, from first principles, that d dθ (cos θ) = -sin θ You may assume the formula for cos(A ± B) and that as h → 0, sin(h)/h → 1 and (cosh - 1)/h → 0 - Edexcel - A-Level Maths Pure - Question 11 - 2018 - Paper 2

Using the cosine addition formula:

$\cos(\theta + h) = \cos \theta \cos h - \sin \theta \sin h$

we substitute this into our limit:

$\frac{d}{d\theta}(\cos \theta) = \lim_{h \to 0} \frac{(\cos \theta \cos h - \sin \theta \sin h) - \cos \theta}{h}$

Next, this simplifies to:

$\frac{d}{d\theta}(\cos \theta) = \lim_{h \to 0} \frac{\cos \theta (\cos h - 1) - \sin \theta \sin h}{h}$

This expression can be split into two parts:

$= \cos \theta \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin \theta \lim_{h \to 0} \frac{\sin h}{h}$

Using the known limits:

$\lim_{h \to 0} \frac{\sin h}{h} = 1$ 2. $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$,

we evaluate:

$\frac{d}{d\theta}(\cos \theta) = \cos \theta (0) - \sin \theta (1)$

which simplifies to:

$\frac{d}{d\theta}(\cos \theta) = -\sin \theta$

Thus, we have shown that:

$\frac{d}{d\theta}(\cos \theta) = -\sin \theta$

This concludes the proof from first principles.