Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 19

\[1)\ y = \sin x + \cos x\]

\[y(x + T) = y(x);\]

\[\sin(x + T) + \cos(x + T) =\]

\[= \sin x + \cos x\]

\[T_{1} = 2\pi;\]

\[T_{2} = 2\pi.\]

\[Ответ:\ \ 2\pi.\]

\[2)\ y = \sin x + tg\ x\]

\[y(x + T) = y(x);\]

\[\sin(x + T) + tg(x + T) =\]

\[= \sin x + tg\ x\]

\[T_{1} = 2\pi;\ \ \ T_{2} = \pi.\]

\[Ответ:\ \ 2\pi.\]

\[3)\ y = \sin x \bullet \sin{3x}\]

\[y = \frac{1}{2}\left( \cos{2x} - \cos{4x} \right)\]

\[y(x + T) = y(x);\]

\[\frac{\cos(2x + 2T) - \cos(4x + 4T)}{2} =\]

\[= \frac{\cos{2x} - \cos{4x}}{2}\]

\[\cos(2x + 2T) - \cos(4x + 4T) =\]

\[= \cos{2x} - \cos{4x}\]

\[2T_{1} = 2\pi;\ \ \ 4T_{2} = 2\pi;\]

\[T_{1} = \pi;\ \ \ \ \ \ \ \ \text{\ \ \ }T_{2} = \frac{\pi}{2}.\]

\[Ответ:\ \ \pi.\]

\[4)\ y = 2\ tg\frac{x}{2} - 3\ tg\frac{x}{3}\]

\[y(x + T) = y(x);\]

\[2\ tg\left( \frac{x}{2} + \frac{T}{2} \right) - 3\ tg\left( \frac{x}{3} + \frac{T}{3} \right) =\]

\[= 2\ tg\frac{x}{2} - 3\ tg\frac{x}{3}\]

\[\frac{T_{1}}{2} = \pi;\ \ \ \text{\ \ \ \ }\frac{T_{2}}{3} = \pi;\]

\[T_{1} = 2\pi;\ \text{\ \ \ }T_{2} = 3\pi.\]

\[Ответ:\ \ 6\pi.\]