For some \( a, b \), let \( f(x) = \left| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} \right| \), where \( x \neq 0 \), \( \lim_{x \to 0} f(x) = \lambda + \mu a + \nu b \).
Then \( (\lambda + \mu + \nu)^2 \) is equal to:
First, compute the determinant of the matrix as \( x \to 0 \) and then take the limit to find the value of \( \lambda + \mu + \nu \). The limit and determinant calculation gives the value 3 for \( \lambda + \mu + \nu \), so squaring this gives 9.
Final Answer: \( (\lambda + \mu + \nu)^2 = 9 \).
Let \( y = y(x) \) be the solution of the differential equation \[ 2\cos x \frac{dy}{dx} = \sin 2x - 4y \sin x, \quad x \in \left( 0, \frac{\pi}{2} \right). \] \( y\left( \frac{\pi}{3} \right) = 0 \), then \( y\left( \frac{\pi}{4} \right) + y\left( \frac{\pi}{4} \right) \) is equal to ________.