Simplify \( \frac{\sin 3 x+\sin x}{\cos 2 x+1} \) to a single trigonometric ratio.

To simplify \( \frac{\sin 3 x + \sin x}{\cos 2 x + 1} \), we start by using the sum-to-product identities for the numerator. The identity states that: \[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \] Applying this to \( \sin 3x + \sin x \): - Let \( A = 3x \) and \( B = x \). Now, we calculate: \[ \sin 3x + \sin x = 2 \sin\left(\frac{3x + x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin(2x) \cos(x) \] Next, we simplify the denominator \( \cos 2x + 1 \). Using the double angle identity for cosine, we have: \[ \cos 2x = 2\cos^2 x - 1 \] Thus, \[ \cos 2x + 1 = 2\cos^2 x \] Now we can substitute both simplified parts into the original expression: \[ \frac{\sin 3 x + \sin x}{\cos 2 x + 1} = \frac{2 \sin(2x) \cos(x)}{2\cos^2 x} \] The \( 2 \) cancels out: \[ = \frac{\sin(2x) \cos(x)}{\cos^2 x} \] We can rewrite this as: \[ = \tan(2x) \sec(x) \] Thus, the final simplified form is: \[ \boxed{\tan(2x) \sec(x)} \]