ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 702

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\[1)\ y = \cos x - 1;\]

\[y(x + 2\pi) = \cos(x + 2\pi) - 1 =\]

\[= \cos x - 1 = y(x);\]

\[2)\ y = \sin x + 1;\]

\[y(x + 2\pi) = \sin(x + 2\pi) + 1 =\]

\[= \sin x + 1 = y(x);\]

\[3)\ y = 3\sin x;\]

\[y(x + 2\pi) = 3\sin(x + 2\pi) =\]

\[= 3\sin x = y(x);\]

\[4)\ y = \frac{\cos x}{2};\]

\[y(x + 2\pi) = \frac{\cos(x + 2\pi)}{2} =\]

\[= \frac{\cos x}{2} = y(x);\]

\[5)\ y = \sin\left( x - \frac{\pi}{4} \right);\]

\[y(x + 2\pi) = \sin\left( x + 2\pi - \frac{\pi}{4} \right) =\]

\[= \sin\left( 2\pi + \left( x - \frac{\pi}{4} \right) \right) =\]

\[= \sin\left( x - \frac{\pi}{4} \right) = y(x);\]

\[6)\ y = \cos\left( x + \frac{2\pi}{3} \right);\]

\[y(x + 2\pi) = \cos\left( x + 2\pi + \frac{2\pi}{3} \right) =\]

\[= \cos\left( 2\pi + \left( x + \frac{2\pi}{3} \right) \right) =\]

\[= \cos\left( x + \frac{2\pi}{3} \right) = y(x).\]