Function Point \[ y=\cos (3 x) \quad\left(\frac{\pi}{4},-\frac{\sqrt{2}}{2}\right) \] (a) Find an equation of the tangent line to the graph of the function at the given polnt. (Let \( x \) be the Independent variable and \( y \) be the dependent variable.) \( y=\cos (3 x) \)

To find the equation of the tangent line to the function \( y = \cos(3x) \) at the point \( \left( \frac{\pi}{4}, -\frac{\sqrt{2}}{2} \right) \), we first need to determine the derivative of the function, which represents the slope of the tangent line. The derivative is: \[ y' = -3\sin(3x). \] Now evaluate the derivative at \( x = \frac{\pi}{4} \): \[ y'\left( \frac{\pi}{4} \right) = -3\sin\left(3 \cdot \frac{\pi}{4}\right) = -3\sin\left(\frac{3\pi}{4}\right) = -3 \cdot \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}. \] Now that we have the slope, we can use the point-slope form of the line: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) = \left( \frac{\pi}{4}, -\frac{\sqrt{2}}{2} \right) \) and \( m = -\frac{3\sqrt{2}}{2} \). Substituting these values in: \[ y + \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}\left(x - \frac{\pi}{4}\right). \] This gives the equation of the tangent line at the point in question!