Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 431

\[\boxed{\mathbf{431.}}\]

\[1)\ \beta = 3\pi = \pi + 2\pi\]

\[\sin\beta = \sin(\pi + 2\pi) = \sin\pi = 0\]

\[\cos\beta = \cos(\pi + 2\pi) =\]

\[= \cos\pi = - 1\]

\[2)\ \beta = 4\pi = 2\pi + 2\pi\]

\[\sin\beta = \sin(2\pi + 2\pi) =\]

\[= \sin{2\pi} = 0\]

\[\cos\beta = \cos(2\pi + 2\pi) =\]

\[= \cos{2\pi} = 1\]

\[3)\ \beta = 3,5\pi = 1,5\pi + 2\pi =\]

\[= \frac{3\pi}{2} + 2\pi\]

\[\sin\beta = \sin\left( \frac{3\pi}{2} + 2\pi \right) = \sin\frac{3\pi}{2} =\]

\[= - 1\]

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

\[= \cos\frac{3\pi}{2} = 0\]

\[4)\ \beta = \frac{5\pi}{2} = \frac{\pi}{2} + \frac{4\pi}{2} = \frac{\pi}{2} + 2\pi\]

\[\sin\beta = \sin\left( \frac{\pi}{2} + 2\pi \right) = \sin\frac{\pi}{2} = 1\]

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

\[= 0\]

\[5)\ \beta = \pi k,\ где\ k \in Z\]

\[\ k - нечетное\ число:\]

\[\beta = \pi k = \pi + \pi(k - 1) =\]

\[= \pi + 2\pi n,\ где\ n \in Z\]

\[\sin\beta = \sin(\pi + 2\pi n) = \sin\pi = 0\]

\[\cos\beta = \cos(\pi + 2\pi n) =\]

\[= \cos\pi = - 1\]

\[k - четное\ число:\]

\[\beta = \pi k = 2\pi n,\ где\ n \in Z\]

\[\sin\beta = \sin(2\pi n) = \sin{2\pi} = 0\]

\[\cos\beta = \cos(2\pi n) = \cos{2\pi} = 1\]

\[Ответ:\ \sin\beta = 0;\ \]

\[\cos\beta = ( - 1)^{k}.\]

\[6)\ \beta = (2k + 1)\pi = \pi + 2\pi k,\ \]

\[где\ k \in Z\]

\[\sin\beta = \sin(\pi + 2\pi k) =\]

\[= \sin\pi = 0\]

\[\cos\beta = \cos(\pi + 2\pi k) = \cos\pi =\]

\[= - 1\]