Given that $\theta$ is measured in radians, prove, from first principles, that $$\frac{d}{d\theta}(\cos \theta) = -\sin \theta$$ 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 $$ - Edexcel - A-Level Maths Pure - Question 10 - 2018 - Paper 2

To prove (\frac{d}{d\theta}(\cos \theta) = -\sin \theta) from first principles, we start by using the definition of the derivative:

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

Using the cosine addition formula, we can rewrite this as:

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

Substituting this into our limit, we get:

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

This simplifies to:

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

Now, separate the limit 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 limits given:

• As ( h \to 0 ), ( \frac{\sin h}{h} \rightarrow 1 )
• As ( h \to 0 ), ( \frac{\cos h - 1}{h} \rightarrow 0 )

We substitute these values into our limit, resulting in:

$= \cos \theta \cdot 0 - \sin \theta \cdot 1$ $= -\sin \theta$

Thus, we conclude that:

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