Task A-3.4. (4 points)

Determine the minimum value of the expression

$$
\sin (x+3)-\sin (x+1)-2 \cos (x+2)
$$

for $x \in \mathbb{R}$.
Solution. We have

$$
\begin{aligned}
& \sin ((x+2)+1)=\sin (x+2) \cos 1+\cos (x+2) \sin 1 \\
& \sin ((x+2)-1)=\sin (x+2) \cos 1-\cos (x+2) \sin 1
\end{aligned}
$$

so by subtracting we get

$$
\sin (x+3)-\sin (x+1)=2 \cos (x+2) \sin 1
$$

Therefore, the expression under consideration is equal to

$$
2 \cos (x+2) \sin 1 - 2 \cos (x+2) = 2 \cos (x+2)(\sin 1 - 1)
$$

Since $2(\sin 1 - 1)$ is a negative real constant, the expression under consideration will be minimal precisely when $\cos (x+2)$ is maximal.

The maximum value of $\cos (x+2)$ is 1,

so the minimum value of the expression under consideration is $2(\sin 1 - 1)$.