ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 538

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\[1)\cos{105{^\circ}} + \cos{75{^\circ}} =\]

\[= 2 \bullet \cos\frac{180{^\circ}}{2} \bullet \cos\frac{30{^\circ}}{2} =\]

\[= 2 \bullet \cos{90{^\circ}} \bullet \cos{15{^\circ}} =\]

\[= 2 \bullet 0 \bullet \cos{15{^\circ}} = 0\]

\[2)\sin{105{^\circ}} - \sin{75{^\circ}} =\]

\[= 2 \bullet \sin\frac{30{^\circ}}{2} \bullet \cos\frac{180{^\circ}}{2} =\]

\[= 2 \bullet \sin{15{^\circ}} \bullet \cos{90{^\circ}} =\]

\[= 2 \bullet \sin{15{^\circ}} \bullet 0 = 0\]

\[3)\cos\frac{11\pi}{12} + \cos\frac{5\pi}{12} =\]

\[= 2 \bullet \cos\frac{16\pi}{24} \bullet \cos\frac{6\pi}{24} =\]

\[= 2 \bullet \cos\frac{2\pi}{3} \bullet \cos\frac{\pi}{4} =\]

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

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

\[= 2 \bullet \left( - \frac{1}{2} \right) \bullet \frac{\sqrt{2}}{2} = - \frac{\sqrt{2}}{2}\]

\[4)\cos\frac{11\pi}{12} - \cos\frac{5\pi}{12} =\]

\[= - 2 \bullet \sin\frac{16\pi}{24} \bullet \sin\frac{6\pi}{24} =\]

\[= - 2 \bullet \sin\frac{2\pi}{3} \bullet \sin\frac{\pi}{4} =\]

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

\[= - 2 \bullet \sin\frac{\pi}{3} \bullet \sin\frac{\pi}{4} =\]

\[= - 2 \bullet \frac{\sqrt{3}}{2} \bullet \frac{\sqrt{2}}{2} = - \frac{\sqrt{6}}{2}\]

\[5)\sin\frac{7\pi}{12} - \sin\frac{\pi}{12} =\]

\[= 2 \bullet \sin\frac{\frac{7\pi}{12} - \frac{\pi}{12}}{2} \bullet \cos\frac{\frac{7\pi}{12} + \frac{\pi}{12}}{2} =\]

\[= 2 \bullet \sin\frac{6\pi}{24} \bullet \cos\frac{8\pi}{24} =\]

\[= 2 \bullet \sin\frac{\pi}{4} \bullet \cos\frac{\pi}{3} = 2 \bullet \frac{\sqrt{2}}{2} \bullet \frac{1}{2} =\]

\[= \frac{\sqrt{2}}{2}\]

\[6)\sin{105{^\circ}} + \sin{165{^\circ}} =\]

\[= 2 \bullet \sin\frac{270{^\circ}}{2} \bullet \cos\left( - \frac{60{^\circ}}{2} \right) =\]

\[= 2 \bullet \sin{135{^\circ}} \bullet \cos( - 30{^\circ}) =\]

\[= 2 \bullet \sin(90{^\circ} + 45{^\circ}) \bullet \cos{30{^\circ}} =\]

\[= 2 \bullet \cos{45{^\circ}} \bullet \cos{30{^\circ}} =\]

\[= 2 \bullet \frac{\sqrt{2}}{2} \bullet \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}\]