Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1100

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\[1)\cos{22{^\circ}} + \cos{24{^\circ}} + \cos{26{^\circ}} +\]

\[+ \cos{28{^\circ}} =\]

\[= 2 \bullet \cos\frac{22{^\circ} + 24{^\circ}}{2} \bullet\]

\[\bullet \cos\frac{22{^\circ} - 24{^\circ}}{2} +\]

\[+ 2 \bullet \cos\frac{26{^\circ} + 28{^\circ}}{2} \bullet\]

\[\bullet \cos\frac{26{^\circ} - 28{^\circ}}{2} =\]

\[= 2 \bullet \cos\frac{46{^\circ}}{2} \bullet \cos\left( - \frac{2{^\circ}}{2} \right) + 2 \bullet\]

\[\bullet \cos\frac{54{^\circ}}{2} \bullet \cos\left( - \frac{2{^\circ}}{2} \right) =\]

\[= 2 \bullet \cos{23{^\circ}} \bullet \cos{1{^\circ}} + 2 \bullet\]

\[\bullet \cos{27{^\circ}} \bullet \cos{1{^\circ}} = 2\cos{1{^\circ}} \bullet\]

\[\bullet \left( \cos{23{^\circ}} + \cos{27{^\circ}} \right) =\]

\[= 2\cos{1{^\circ}} \bullet 2 \bullet \cos\frac{23{^\circ} + 27{^\circ}}{2} \bullet\]

\[\bullet \cos\frac{23{^\circ} - 27{^\circ}}{2} = 4\cos{1{^\circ}} \bullet\]

\[\bullet \cos\frac{50{^\circ}}{2} \bullet \cos\left( - \frac{4{^\circ}}{2} \right) =\]

\[= 4\cos{1{^\circ}} \bullet \cos{2{^\circ}} \bullet \cos{25{^\circ}}\]

\[2)\cos\frac{\pi}{12} + \cos\frac{\pi}{4} + \cos\frac{5\pi}{6} =\]

\[= 2 \bullet \cos\frac{\frac{\pi}{12} + \frac{\pi}{4}}{2} \bullet \cos\frac{\frac{\pi}{12} - \frac{\pi}{4}}{2} +\]

\[+ \cos\left( \pi - \frac{\pi}{6} \right) =\]

\[= 2 \bullet \cos\frac{4\pi}{24} \bullet \cos\left( - \frac{2\pi}{24} \right) -\]

\[- \cos\frac{\pi}{6} = 2 \bullet \cos\frac{\pi}{6} \bullet \cos\frac{\pi}{12} -\]

\[- \cos\frac{\pi}{6} =\]

\[= 2\cos\frac{\pi}{6} \bullet \left( \cos\frac{\pi}{12} - \frac{1}{2} \right) =\]

\[= 2 \bullet \frac{\sqrt{3}}{2} \bullet \left( \cos\frac{\pi}{12} - \cos\frac{\pi}{3} \right) =\]

\[= \sqrt{3} \bullet ( - 2) \bullet \sin\frac{\frac{\pi}{12} + \frac{\pi}{3}}{2} \bullet\]

\[\bullet \sin\frac{\frac{\pi}{12} - \frac{\pi}{3}}{2} = - 2\sqrt{3} \bullet \sin\frac{5\pi}{24} \bullet\]

\[\bullet \sin\left( - \frac{3\pi}{24} \right) =\]

\[= 2\sqrt{3} \bullet \sin\frac{5\pi}{24} \bullet \sin\frac{\pi}{8}\]