ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1099

\[\boxed{\mathbf{1099}\mathbf{.}}\]

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

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

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

\[2 \bullet \cos\frac{2a + \frac{\pi}{2} - 2a}{2} \bullet\]

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

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

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

\[\sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right) =\]

\[= \sqrt{2}\cos\left( 2a - \frac{\pi}{4} \right)\]

\[Что\ и\ требовалось\ доказать.\]

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

\[+ \cos\left( \frac{2\pi}{3} - a \right) = 0\]

\[\cos a + 2 \bullet \cos\frac{\frac{2\pi}{3} + a + \frac{2\pi}{3} - a}{2} \bullet\]

\[\bullet \cos\frac{\frac{2\pi}{3} + a - \frac{2\pi}{3} + a}{2} = 0\]

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

\[\cos a + 2 \bullet \cos\left( \pi - \frac{\pi}{3} \right) \bullet \cos a = 0\]

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

\[\cos a - 2 \bullet \frac{1}{2} \bullet \cos a = 0\]

\[\cos a - \cos a = 0\]

\[0 = 0\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ \frac{\sin{2a} + \sin{5a} - \sin{3a}}{\cos a + 1 - 2\sin^{2}{2a}} =\]

\[= 2\sin a\]