f(x) = sin x Using differentiation from first principles find the exact value of \( f \left( \frac{\pi}{6} \right) \) Fully justify your answer. - AQA - A-Level Maths Mechanics - Question 17 - 2017 - Paper 1

Next, we apply the sum of angles identity for sine:

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

Thus, we can write:

$\sin \left( \frac{\pi}{6} + h \right) = \sin \left( \frac{\pi}{6} \right) \cos(h) + \cos \left( \frac{\pi}{6} \right) \sin(h)$

This gives us:

$= \lim_{h \to 0} \frac{\left( \sin \left( \frac{\pi}{6} \right) \cos h + \cos \left( \frac{\pi}{6} \right) \sin h \right) - \sin \left( \frac{\pi}{6} \right)}{h}$

Rearranging the expression, we can combine the terms:

$= \lim_{h \to 0} \frac{\sin \left( \frac{\pi}{6} \right) (\cos h - 1) + \cos \left( \frac{\pi}{6} \right) \sin h}{h}$

Breaking it down into two separate limits:

$= \lim_{h \to 0} \left( \sin \left( \frac{\pi}{6} \right) \frac{\cos h - 1}{h} + \cos \left( \frac{\pi}{6} \right) \frac{\sin h}{h} \right)$

Using known limits:

1. (\lim_{h \to 0} \frac{\sin h}{h} = 1)
2. (\lim_{h \to 0} \frac{\cos h - 1}{h} = 0)

We can now substitute these limits into our previous expression:

$= \sin \left( \frac{\pi}{6} \right) \cdot 0 + \cos \left( \frac{\pi}{6} \right) \cdot 1$

Given that (\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}) and (\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}):

Thus,

$f' \left( \frac{\pi}{6} \right) = 0 + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

Therefore, the final exact value is (f' \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}).