f(x) = sin.x Using differentiation from first principles find the exact value of $rac{ ext{ }rac{ ho}{6}}{6}$ Fully justify your answer. - AQA - A-Level Maths Pure - Question 17 - 2017 - Paper 1

Next, we apply the sine addition formula, which states that:

$\text{sin}(A + B) = \text{sin}(A)\text{cos}(B) + \text{cos}(A)\text{sin}(B)$

We can rewrite the expression:

$f'\left(\frac{\pi}{6}\right) = \text{lim}_{h \to 0} \frac{\text{sin}\left(\frac{\pi}{6}\right)\text{cos}(h) + \text{cos}\left(\frac{\pi}{6}\right)\text{sin}(h) - \text{sin}\left(\frac{\pi}{6}\right)}{h}$

This simplifies to:

$f'\left(\frac{\pi}{6}\right) = \text{lim}_{h \to 0} \frac{\text{sin}\left(\frac{\pi}{6}\right)\text{cos}(h) - \text{sin}\left(\frac{\pi}{6}\right) + \text{cos}\left(\frac{\pi}{6}\right)\text{sin}(h)}{h}$

Factoring out $\text{sin}\left(\frac{\pi}{6}\right)$ gives us:

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

From the limit properties, we know:

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

Thus, substituting back in gives:

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

We know that:

$\text{cos}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

Thus, the exact value of $f'\left(\frac{\pi}{6}\right)$ is:

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