Given that θ is measured in radians, prove, from first principles, that the derivative of sin θ is cos θ - Edexcel - A-Level Maths Pure - Question 11 - 2017 - Paper 1

Now, simplifying the expression:

$\frac{\sin θ \cos h - \sin θ + \cos θ \sin h}{h} = \frac{\sin θ (\cos h - 1)}{h} + \cos θ \frac{\sin h}{h}$

As h approaches 0,

• We know that ( \frac{\sin h}{h} \rightarrow 1 ),
• ( \cos h - 1 \rightarrow 0 ) leading to ( \frac{\cos h - 1}{h} \rightarrow 0 ).

Thus, the limit of the overall expression is:

$\lim_{h \rightarrow 0} \left( \frac{\sin θ (\cos h - 1)}{h} + \cos θ \frac{\sin h}{h} \right) = 0 + \cos θ \cdot 1$

Therefore, we conclude that:

$\frac{d}{dθ} \sin θ = \cos θ$.

This proves that the derivative of sin θ is indeed cos θ.