Example 8. Find the Wronskian determinant for the functions: $y_{1}(x)=\sin x$,

$$
y_{2}(x)=\sin \left(x+\frac{\pi}{8}\right), y_{3}(x)=\sin \left(x-\frac{\pi}{8}\right)
$$
Solution. We have

$$
W\left[y_{1}, y_{2}, y_{3}\right]=\left|\begin{array}{rrr}
\sin x & \sin \left(x+\frac{\pi}{8}\right) & \sin \left(x-\frac{\pi}{8}\right) \\
\cos x & \cos \left(x+\frac{\pi}{8}\right) & \cos \left(x-\frac{\pi}{8}\right) \\
-\sin x & -\sin \left(x+\frac{\pi}{8}\right) & -\sin \left(x-\frac{\pi}{8}\right)
\end{array}\right|=0
$$

since the first and last rows of the determinant are proportional.

Theorem. If the system of functions $y_{1}(x), y_{2}(x), \ldots, y_{n}(x)$ is linearly dependent on the interval $[a, b]$, then its Wronskian determinant is identically zero on this interval.

For example, the system of functions $\sin x, \sin \left(x+\frac{\pi}{8}\right), \sin \left(x-\frac{\pi}{8}\right)$ is linearly dependent in the interval $(-\infty,+\infty)$, and the Wronskian determinant of these functions is zero everywhere in this interval (see examples 4 and 8).

This theorem provides a necessary condition for the linear dependence of a system of functions. The converse statement is false, i.e., the Wronskian determinant can be identically zero even when the given functions form a linearly independent system on some interval.