Find the equation of the tangent to the curve x = cos(2y + π) at \( \left( 0, \frac{\pi}{4} \right) \) - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 2

Using the chain rule, we find dy/dx:

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

Substituting ( y = \frac{\pi}{4} ) into the expression:

$\frac{dy}{dx} = \frac{1}{-2\sin(\frac{\pi}{2})} = \frac{1}{-2} = -\frac{1}{2}$

Using the point-slope formula for the line, y - y_1 = m(x - x_1):

Substituting the slope ( m = -\frac{1}{2} ) and the point ( \left( 0, \frac{\pi}{4} \right) ):

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

This simplifies to:

[ y = -\frac{1}{2}x + \frac{\pi}{4} ]