ГДЗ по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 32

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\[\textbf{а)}\ y = \sin^{2}x = \frac{1 - \cos{2x}}{2}\]

\[y = \cos{2x}:\]

\[T = \frac{2\pi}{2} = \pi.\]

\[Периодическая;\ \ T = \pi.\]

\[\textbf{б)}\ y = \cos^{2}x = \frac{1 + \cos{2x}}{2}\]

\[y = \cos{2x}:\]

\[T = \frac{2\pi}{2} = \pi.\]

\[Периодическая;\ \ T = \pi.\]

\[\textbf{в)}\ y = \sin^{2}x - \cos^{2}x =\]

\[= - \left( \cos^{2}x - \sin^{2}x \right) = - \cos{2x}\]

\[T = \frac{2\pi}{2} = \pi.\]

\[Периодическая;\ \ T = \pi.\]

\[\textbf{г)}\ y = 1 + tg\ x\]

\[\text{tg}(x - \pi) = tg\ x = tg\ (x + \pi)\]

\[T = \pi.\]

\[Периодическая;\ \ T = \pi.\]

\[\textbf{д)}\ y = 1 + ctg\ x\]

\[\text{ctg\ }(x - \pi) = ctg\ x = ctg\ (x + \pi)\]

\[T = \pi.\]

\[Периодическая;\ \ T = \pi.\]

\[\textbf{е)}\ y = \sin\sqrt{x};\ \ X \geq 0\]

\[Не\ периодическая.\]

\[\textbf{ж)}\ y = \cos\sqrt{- x};\ \ X \leq 0\]

\[Не\ периодическая.\]

\[\textbf{з)}\ y = tg\ \sqrt{x};\ \ X \geq 0\]

\[Не\ периодическая.\]

\[\textbf{и)}\ y = ctg\ \sqrt{- x};\ \ X \leq 0\]

\[Не\ периодическая.\]