Find the equation of the tangent to the curve $x = ext{cos}(2y + heta)$ at \( \left( 0, \frac{\pi}{4} \right) \) - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 2

Next, we apply the chain rule to find the derivative $abla = \frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{1}{2 \sin(2y + \theta)}$

Substituting $y = \frac{\pi}{4}$ into our derivative:

$\frac{dy}{dx} \bigg|_{y = \frac{\pi}{4}} = \frac{1}{2 \sin(\frac{\pi}{2} + \theta)} = \frac{1}{2}$

Now, using the point-slope form of a line:

$y - \frac{\pi}{4} = \frac{1}{2} (x - 0)$

Rearranging gives:

$y = \frac{1}{2}x + \frac{\pi}{4}$

Thus, we find: $a = \frac{1}{2}, b = \frac{\pi}{4}$