Using small angle approximations, show that for small, non-zero values of $x$: $$\frac{x \tan 5x}{\cos 4x - 1} \approx A$$ where $A$ is a constant to be determined. - AQA - A-Level Maths Pure - Question 4 - 2020 - Paper 2

The cosine function can be approximated as:

$\cos \theta \approx 1 - \frac{\theta^2}{2}$

Thus for (\cos 4x):

$\cos 4x \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2$

Substituting the approximations into the expression, we have:

$\frac{x \tan 5x}{\cos 4x - 1} \approx \frac{x (5x)}{(1 - 8x^2) - 1}$

This simplifies to:

$\frac{5x^2}{-8x^2} = -\frac{5}{8}$

From the previous step, we find:

$A = -\frac{5}{8}$

Thus, we have shown that:

$\frac{x \tan 5x}{\cos 4x - 1} \approx -\frac{5}{8}$