Answer
$f$ satisfies the wave equation and $f$ can be represented as the sum of a wave traveling to the left and a wave traveling to the right as
$$f(z, t) = \frac{A}{2} [\sin(kz+kvt)+\sin(kz-kvt)] $$
Work Step by Step
Wave equation is given by:
$$ \frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} $$
For $f(z, t) = A \sin(kz)\cos(kvt) $
$$ \frac{\partial f}{\partial z} = Ak\cos(kz)cos(kvt) $$
$$ \frac{\partial^2 f}{\partial z^2} = -Ak^2\sin(kz)\cos(kvt)\qquad ...(1)$$
$$ \frac{\partial f}{\partial t} = -Akv\sin(kz)\sin(kvt) $$
$$ \frac{\partial^2 f}{\partial t^2} = -Ak^2v^2\sin(kz)\cos(kvt)\qquad ...(2)$$
From equation $(1) \text{ and } (2)$,
$$ \frac{\partial^2 f_1}{\partial z^2} = \frac{1}{v^2}\frac{\partial^2 f_1}{\partial t^2} $$
Hence $f$ satisfies the wave equation.
Using the Trigonometry Identity,
$$\sin(A)\cos(B) = \frac{1}{2}[\sin(A+B)+\sin(A-b)]$$
we get,
$$f(z, t) = A \sin(kz)\cos(kvt) $$
$$f(z, t) = \frac{A}{2} [\sin(kz+kvt)+\sin(kz-kvt)] $$
Hence, $f$ can be represented as the sum of a wave traveling to the left and a wave traveling to the right.
