Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1319

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\[1)\ \frac{1 - 2\sin^{2}a}{1 + \sin{2a}} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{1 - 2\sin^{2}a}{1 + \sin{2a}} =\]

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

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

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

\[= \frac{\cos a - \sin a}{\cos a + \sin a} = \frac{\frac{\cos a}{\cos a} - \frac{\sin a}{\cos a}}{\frac{\cos a}{\cos a} + \frac{\sin a}{\cos a}} =\]

\[= \frac{1 - tg\ a}{1 + tg\ a}.\]

\[Тождество\ доказано.\]

\[2)\ \frac{1}{4\sin^{2}a \bullet \cos^{2}a} = 1 + \frac{\left( 1 - tg^{2}\text{\ a} \right)^{2}}{4\ tg^{2}\text{\ a}}\]

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

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

\[= ctg^{2}\ 2a + 1;\]

\[1 + \frac{\left( 1 - tg^{2}\text{\ a} \right)^{2}}{4\ tg^{2}\text{\ a}} =\]

\[= 1 + \left( \frac{1 - tg^{2}\text{\ a}}{2\ tg\ a} \right)^{2} =\]

\[= 1 + \left( \frac{1}{tg\ 2a} \right)^{2} = 1 + ctg^{2}\ 2a.\]

\[Тождество\ доказано.\]

\[3)\ tg\left( \frac{\pi}{4} + a \right) = \frac{1 + \sin{2a}}{\cos{2a}}\]

\[\text{tg}\left( \frac{\pi}{4} + a \right) = \frac{\sin\left( \frac{\pi}{4} + a \right)}{\cos\left( \frac{\pi}{4} + a \right)} =\]

\[= \frac{\sin\frac{\pi}{4} \bullet \cos a + \cos\frac{\pi}{4} \bullet \sin a}{\cos\frac{\pi}{4} \bullet \cos a - \sin\frac{\pi}{4} \bullet \sin a} =\]

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

\[= \frac{\cos a + \sin a}{\cos a - \sin a} =\]

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

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

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

\[Тождество\ доказано.\]

\[4)\ \frac{1 - \sin{2a}}{1 + \sin{2a}} = ctg^{2}\left( \frac{\pi}{4} + a \right)\]

\[\text{ct}g^{2}\left( \frac{\pi}{4} + a \right) = \frac{\cos^{2}\left( \frac{\pi}{4} + a \right)}{\sin^{2}\left( \frac{\pi}{4} + a \right)} =\]

\[= \frac{\left( \cos\frac{\pi}{4} \bullet \cos a - \sin\frac{\pi}{4} \bullet \sin a \right)^{2}}{\left( \sin\frac{\pi}{4} \bullet \cos a + \cos\frac{\pi}{4} \bullet \sin a \right)^{2}} =\]

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

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

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

\[= \frac{1 - \sin{2a}}{1 + \sin{2a}}.\]

\[Тождество\ доказано.\]