Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 518

\[\boxed{\mathbf{518}\mathbf{.}}\]

\[1)\ \frac{1 - \cos a}{\sin a} =\]

\[= \frac{2}{\sin a} \bullet \frac{1 - \cos a}{2} =\]

\[= \frac{2}{2\sin\frac{a}{2} \bullet \cos\frac{a}{2}} \bullet \sin^{2}\frac{a}{2} =\]

\[= \frac{\sin\frac{a}{2}}{\cos\frac{a}{2}} = \ tg\frac{a}{2}\]

\[2)\ \frac{\sin a}{1 + \cos a} =\]

\[= \sin a \bullet \frac{1}{2} \bullet \left( \frac{1 + \cos a}{2} \right)^{- 1} =\]

\[= \frac{2\sin\frac{a}{2} \bullet \cos\frac{a}{2}}{2} \bullet \left( \cos^{2}\frac{a}{2} \right)^{- 1} =\]

\[= \frac{\sin\frac{a}{2} \bullet \cos\frac{a}{2}}{\cos^{2}\frac{a}{2}} = \frac{\sin\frac{a}{2}}{\cos\frac{a}{2}} = tg\frac{a}{2}\]

\[3)\ \frac{1 - \cos{2a} + \sin{2a}}{1 + \cos{2a} + \sin{2a}} =\]

\[= \frac{2\sin^{2}a + 2\sin a \bullet \cos a}{2\cos^{2}a + 2\sin a \bullet \cos a} =\]

\[= \frac{2\sin a\left( \sin a + \cos a \right)}{2\cos a\left( \cos a + \sin a \right)} =\]

\[= \frac{\sin a}{\cos a} = tg\ a\]

\[4)\ \frac{1 + \cos{4a}}{\sin{4a}} =\]

\[= \frac{2}{\sin{4a}} \bullet \frac{1 + \cos{4a}}{2} =\]

\[= \frac{2}{2\sin{2a} \bullet \cos{2a}} \bullet \cos^{2}{2a} =\]

\[= \frac{\cos{2a}}{\sin{2a}} = ctg\ 2a\]

\[5)\ \frac{1 + \cos{2a} + \sin{2a}}{\sin a + \cos a} =\]

\[= \frac{2\cos^{2}a + 2\sin a \bullet \cos a}{\sin a + \cos a} =\]

\[= \frac{2\cos a\left( \cos a + \sin a \right)}{\sin a + \cos a} =\]

\[= 2\cos a\]

\[6)\ \left( 1 - \cos{2a} \right) \bullet ctg\ a =\]

\[= 2\sin^{2}a \bullet \frac{\cos a}{\sin a} =\]

\[= 2\sin a \bullet \cos a = \sin{2a}\]