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

Now, substitute $y = \frac{\pi}{4}$:

1. Compute $\frac{dx}{dy}$ at this point: $\frac{dx}{dy} = -2\sin(2(\frac{\pi}{4}) + \pi) = -2\sin(\frac{\pi}{2} + \pi) = -2\sin(\frac{3\pi}{2}) = 2$
2. Therefore, we find: $\frac{dy}{dx} = \frac{1}{2}$

With the slope $\frac{dy}{dx} = \frac{1}{2}$ at the point $\left(0, \frac{\pi}{4}\right)$, we can now use the point-slope form of the line:

1. Write the equation: $y - y_1 = m(x - x_1)$ where $m = \frac{1}{2}$, $x_1 = 0$, and $y_1 = \frac{\pi}{4}$.
2. Plugging in the values gives: $y - \frac{\pi}{4} = \frac{1}{2}(x - 0)$ Simplifying this: $y = \frac{1}{2}x + \frac{\pi}{4}$