Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 506

\[\boxed{\mathbf{506}\mathbf{.}}\]

\[1)\ 2\cos{40{^\circ}} \bullet \cos{50{^\circ}} =\]

\[= 2 \bullet \cos(90{^\circ} - 50{^\circ}) \bullet \cos{50{^\circ}} =\]

\[= 2\sin{50{^\circ}} \bullet \cos{50{^\circ}} =\]

\[= \sin(2 \bullet 50{^\circ}) = \sin{100{^\circ}} =\]

\[= \sin{80{^\circ}}\]

\[2)\ 2\sin{25{^\circ}} \bullet \sin{65{^\circ}} =\]

\[= 2 \bullet \cos(90{^\circ} - 65{^\circ}) \bullet \sin{65{^\circ}} =\]

\[= 2\cos{65{^\circ}} \bullet \sin{65{^\circ}} =\]

\[= \sin(2 \bullet 65{^\circ}) = \sin{130{^\circ}} =\]

\[= \sin{50{^\circ}}\]

\[3)\sin{2a} + \left( \sin a - \cos a \right)^{2} =\]

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

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

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

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

\[= \cos^{2}{2a} - \sin^{2}{2a} + \sin^{2}\text{ax} =\]

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