ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1121

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\[1)\cos\frac{23\pi}{4} - \sin\frac{15\pi}{4} =\]

\[= \cos\left( 6\pi - \frac{\pi}{4} \right) - \sin\left( 4\pi - \frac{\pi}{4} \right) =\]

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

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

\[2)\sin\frac{25\pi}{3} - tg\frac{10\pi}{3} =\]

\[= \sin\left( 8\pi + \frac{\pi}{3} \right) - tg\left( 3\pi + \frac{\pi}{3} \right) =\]

\[= \sin\frac{\pi}{3} - tg\frac{\pi}{3} = \frac{\sqrt{3}}{2} - \sqrt{3} =\]

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

\[3)\ 3\cos{3660{^\circ}} + \sin( - 1560{^\circ}) =\]

\[= 3\cos(10 \bullet 360{^\circ} + 60{^\circ}) +\]

\[+ \sin( - 5 \bullet 360{^\circ} + 240{^\circ}) =\]

\[= 3\cos{60{^\circ}} + \sin{240{^\circ}} =\]

\[= 3 \bullet \frac{1}{2} + \sin(180{^\circ} + 60{^\circ}) = \frac{3}{2} -\]

\[- \sin{60{^\circ}} = \frac{3}{2} - \frac{\sqrt{3}}{2} = \frac{3 - \sqrt{3}}{2}\]

\[4)\cos( - 945{^\circ}) + tg\ 1035{^\circ} =\]

\[= \cos( - 3 \bullet 360{^\circ} + 135{^\circ}) +\]

\[+ \text{tg}(5 \bullet 180{^\circ} + 135{^\circ}) =\]

\[= \cos{135{^\circ}} + tg\ 135{^\circ} =\]

\[= \cos(90{^\circ} + 45{^\circ}) +\]

\[+ \text{tg}(90{^\circ} + 45{^\circ}) = - \sin{45{^\circ}} -\]

\[- ctg\ 45{^\circ} = - \frac{\sqrt{2}}{2} - 1 =\]

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