Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1084

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

\[- \text{ctg}\left( - \frac{11\pi}{2} \right) =\]

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

\[- \text{ctg}\left( - 6\pi + \frac{\pi}{2} \right) =\]

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

\[- \text{ctg}\frac{\pi}{2} = \cos\frac{\pi}{4} + \sin\frac{\pi}{4} - 0 =\]

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

\[2)\sin\frac{25\pi}{3} - \cos\left( - \frac{17\pi}{2} \right) -\]

\[- \text{tg}\frac{10\pi}{3} =\]

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

\[- \cos\left( - 8\pi - \frac{\pi}{2} \right) -\]

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

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

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

\[- 0 - \frac{2\sqrt{3}}{2} = - \frac{\sqrt{3}}{2}\]

\[3)\sin( - 7\pi) - 2\cos\frac{31\pi}{3} -\]

\[- \text{tg}\frac{7\pi}{4} =\]

\[= \sin( - 8\pi + \pi) -\]

\[- 2\cos\left( 10\pi + \frac{\pi}{3} \right) -\]

\[- \text{tg}\left( 2\pi - \frac{\pi}{4} \right) =\]

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

\[= 0 - 2 \bullet \frac{1}{2} + tg\frac{\pi}{4} = - 1 + 1 = 0\]

\[4)\cos( - 9\pi) + 2\sin\left( - \frac{49\pi}{6} \right) -\]

\[- \text{ctg}\left( - \frac{21\pi}{4} \right) =\]

\[= \cos( - 10\pi + \pi) +\]

\[+ 2\sin\left( - 8\pi - \frac{\pi}{6} \right) -\]

\[- \text{ctg}\left( - 5\pi - \frac{\pi}{4} \right) =\]

\[= \cos\pi + 2\sin\left( - \frac{\pi}{6} \right) -\]

\[- \text{ctg}\left( - \frac{\pi}{4} \right) = - 1 - 2 \bullet \sin\frac{\pi}{6} +\]

\[+ \text{ctg}\frac{\pi}{4} =\]

\[= - 1 - 2 \bullet \frac{1}{2} + 1 = - 1\]