Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 499

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

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

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

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

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

\[= 2\sin\left( \frac{\pi}{8} + \frac{\beta}{2} \right) \bullet \cos\left( \frac{\pi}{8} + \frac{\beta}{2} \right)\]

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

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

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

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

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

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

\[5)\sin a = 2\sin\frac{a}{2} \bullet \cos\frac{a}{2}\]

\[6)\cos a = \cos^{2}\frac{a}{2} - \sin^{2}\frac{a}{2}\]