ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1061

\[\boxed{\mathbf{1061}\mathbf{.}}\]

\[1)\cos\frac{\pi}{5} \cdot \cos\frac{2\pi}{5} = 0,25\]

\[\sin\frac{\pi}{5} \cdot \cos\frac{\pi}{5} \cdot \frac{\cos\frac{2\pi}{5}}{\sin\frac{\pi}{5}} = 0,25\]

\[\frac{1}{2} \cdot \sin\frac{2\pi}{5} \cdot \frac{\cos\frac{2\pi}{5}}{\sin\frac{\pi}{5}} = 0,25\]

\[\frac{1}{4} \cdot \frac{\sin\frac{4\pi}{5}}{\sin\frac{\pi}{5}} = 0,25\]

\[\frac{1}{4} \cdot \frac{\sin\left( \pi - \frac{\pi}{5} \right)}{\sin\frac{\pi}{5}} = 0,25\]

\[\frac{1}{4} = 0,25\]

\[0,25 = 0,25\]

\[Равенство\ доказано.\]

\[2)\ 8\cos{10{^\circ}} \cdot \cos{20{^\circ}} \cdot \cos{40{^\circ}} =\]

\[= ctg\ 10{^\circ}\ \]

\[8\cos{10{^\circ}} \bullet \cos{20{^\circ}} \bullet \cos{40{^\circ}} =\]

\[= ctg\ 10{^\circ};\]

\[\frac{4}{\sin{10{^\circ}}} \bullet \left( 2\sin{10{^\circ}} \bullet \cos{10{^\circ}} \right) \bullet\]

\[\bullet \cos{20{^\circ}} \bullet \cos{40{^\circ}} = ctg\ 10{^\circ};\]

\[\frac{2}{\sin{10{^\circ}}} \bullet \left( 2\sin{20{^\circ}} \bullet \cos{20{^\circ}} \right) \bullet\]

\[\bullet \cos{40{^\circ}} = ctg\ 10{^\circ};\]

\[\frac{2\sin{40{^\circ}} \bullet \cos{40{^\circ}}}{\sin{10{^\circ}}} = ctg\ 10{^\circ};\]

\[\frac{\sin{80{^\circ}}}{\sin{10{^\circ}}} = ctg\ 10{^\circ};\]

\[\frac{\sin(90{^\circ} - 10{^\circ})}{\sin{10{^\circ}}} = ctg\ 10{^\circ};\]

\[\frac{\cos{10{^\circ}}}{\sin{10{^\circ}}} = ctg\ 10{^\circ};\]

\[ctg\ 10{^\circ} = ctg\ 10{^\circ};\]

\[Равенство\ доказано.\]

\[3)\ 16\cos{20{^\circ}} \cdot \cos{40{^\circ}} \cdot\]

\[\bullet \cos{60{^\circ}} \cdot \cos{80{^\circ}} = 1\]

\[5)\ \frac{\text{tg}\frac{a}{2}}{\cos a} = tg\ a - tg\frac{a}{2}\]

\[tg\ a - tg\frac{a}{2} = \frac{\text{tg}\frac{a}{2}}{\cos a}\]

\[\frac{2\text{tg}\frac{a}{2}}{1 - tg\frac{a}{2}} - tg\frac{a}{2} = \frac{\text{tg}\frac{a}{2}}{\cos a}\]

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

\[\text{tg}\frac{a}{2} \cdot \frac{1 + tg^{2}\frac{a}{2}}{1 - tg^{2}\frac{a}{2}} = \frac{\text{tg}\frac{a}{2}}{\cos a}\]

\[\text{tg}\frac{a}{2} \cdot \frac{\cos^{2}\frac{a}{2} + \sin^{2}\frac{a}{2}}{\cos^{2}\frac{a}{2} - \sin^{2}\frac{a}{2}} = \frac{\text{tg}\frac{a}{2}}{\cos a}\]

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

\[Равенство\ доказано.\]

\[6)\sin{3a}\sin^{3}a +\]

\[+ \cos{3a} \cdot \cos^{3}a = \cos{2a}\]

\[Преобразуем\ левую\ часть\]

\[\ равенства:\]

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

\[\cdot \sin^{2}a +\]

\[+ \left( \cos{2a}\cos a - \sin a\sin{2a} \right) \cdot\]

\[\cdot {cos²}a =\]

\[{= \sin^{4}}a\cos{2a} +\]

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

\[+ \cos{2a}\cos^{4}a -\]

\[- \sin a\cos^{3}a\sin{2a} =\]

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

\[+ \sin{2a}\cos a\sin a \cdot\]

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

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

\[= \cos^{4}a +\]

\[+ 2\sin^{2}a\cos^{2}a + \sin^{4}a\]

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

\[A:\]

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

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

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

\[= \cos{2a} \cdot 1 = \cos{2a}\]

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

\[Равенство\ доказано.\]