The curve C, in the standard Cartesian plane, is defined by the equation $x = 4 \, ext{sin} \, 2y$ for $-\frac{\pi}{4} < y < \frac{\pi}{4}$ - Edexcel - A-Level Maths Pure - Question 2 - 2019 - Paper 2

Using the small angle approximation, we know that for small $y$, $\text{sin}(2y) \approx 2y$. Therefore, substituting this into the original equation yields:

$x = 4 \cdot 2y = 8y \implies y = \frac{x}{8}.$

The answer from (a) gives the slope of the tangent line to the curve at the origin, which is $\frac{1}{8}$. The equation found in (b)(i), $y = \frac{x}{8}$, represents a straight line whose slope is also $\frac{1}{8}$. Thus, both answers are consistent, indicating that the tangent to the curve at the origin matches the line described by the approximation for small angles.

To show this, we start from the initial derivative:

$\frac{dy}{dx} = \frac{1}{8 \cos(2y)}.$

Using the identity $\sin^2(2y) + \cos^2(2y) = 1$, we can substitute $\cos(2y) = \sqrt{1 - \sin^2(2y)}$. Substituting back into our equation:

$\frac{dy}{dx} = \frac{1}{8 \sqrt{1 - \sin^2(2y)}}.$

Since $x = 4\sin(2y)$, we can express $\sin(2y)$ in terms of $x$:

$\sin(2y) = \frac{x}{4},$

therefore:

$\frac{dy}{dx} = \frac{1}{8 \sqrt{1 - \left(\frac{x}{4}\right)^2}} = \frac{1}{8 \sqrt{1 - \frac{x^2}{16}}} = \frac{1}{a \sqrt{b - x^2}}.$

Choosing $a = 2$ and $b = 4$ will satisfy the equation, therefore showing the proof.