Problem 781

11.51 In the coordinate plane $x O y$, let's construct the point $A(5 ; 12)$ and consider the vector $\overline{O A}$ (Fig. 11.18). Suppose this vector forms an angle $\varphi$ with the x-axis. Then, from the figure, it is clear that $\quad \cos \varphi=\frac{5}{\sqrt{25+144}}=\frac{5}{13}, \quad \sin \varphi=\frac{12}{\sqrt{25+144}}=\frac{12}{13}$, $\operatorname{tg} \varphi=\frac{12}{5}, \quad$ i.e., $\varphi=\operatorname{arctg} \frac{12}{5}$. Let's transform the given function as follows:

$$\begin{aligned} & y=13\left(\frac{5}{13} \sin x-\frac{12}{13} \cos x\right)= \\ & =13(\cos \varphi \sin x-\sin \varphi \cos x)=13 \sin (x-\varphi) \end{aligned}$$

where $\varphi=\operatorname{arctg} \frac{12}{5}$. Clearly, $-1 \leq \sin (x-\varphi) \leq 1$, hence $-13 \leq y \leq 13$.

Answer: $-13 \leq y \leq 13$.

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Fig. 11.18