ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1094

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

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

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

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

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

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

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

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

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

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

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

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

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