Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 537

\[\boxed{\mathbf{537}\mathbf{.}}\]

\[1)\sin\left( \frac{\pi}{3} + a \right) + \sin\left( \frac{\pi}{3} - a \right) =\]

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

\[= 2 \bullet \frac{\sqrt{3}}{2} \bullet \cos a = \sqrt{3}\cos a\]

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

\[= - 2 \bullet \sin\frac{\pi}{4} \bullet \sin( - \beta) =\]

\[= 2 \bullet \frac{\sqrt{2}}{2} \bullet \sin\beta = \sqrt{2}\sin\beta\]

\[3)\sin^{2}\left( \frac{\pi}{4} + a \right) - \sin^{2}\left( \frac{\pi}{4} - a \right) =\]

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

\[= \sin{2a}\]

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

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

\[= \sin{2a}\]