ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1083

\[\boxed{\mathbf{1083}\mathbf{.}}\]

\[1)\cos{630{^\circ}} - \sin{1470{^\circ}} -\]

\[- ctg\ 1125{^\circ} =\]

\[= \cos(2 \bullet 360{^\circ} - 90{^\circ}) -\]

\[- \sin(4 \bullet 360{^\circ} + 30{^\circ}) -\]

\[- \text{ctg}(6 \bullet 180{^\circ} + 45{^\circ}) =\]

\[= \cos( - 90{^\circ}) - \sin(30{^\circ}) -\]

\[- ctg\ 45{^\circ} = - \cos{90{^\circ}} - \frac{1}{2} -\]

\[- 1 = 0 - \frac{1}{2} - \frac{2}{2} = - \frac{3}{2}\]

\[2)\ tg\ 1800{^\circ} - \sin{495{^\circ}} +\]

\[+ \cos{945{^\circ}} =\]

\[= tg\ (10 \bullet 180{^\circ} + 0{^\circ}) -\]

\[- \sin(360{^\circ} + 180{^\circ} - 45{^\circ}) +\]

\[+ \cos(2 \bullet 360{^\circ} + 180{^\circ} + 45{^\circ}) =\]

\[= tg\ 0 - \sin(180{^\circ} - 45{^\circ}) +\]

\[+ \cos(180{^\circ} + 45{^\circ}) = 0 -\]

\[- \sin{45{^\circ}} - \cos{45{^\circ}} =\]

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

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

\[+ \cos( - 450{^\circ}) =\]

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

\[+ \sin( - 5 \bullet 360{^\circ} + 180{^\circ} + 60{^\circ}) +\]

\[\text{+}\cos( - 360{^\circ} - 90{^\circ}) =\]

\[= 3\cos{60{^\circ}} + \sin(180{^\circ} + 60{^\circ}) +\]

\[+ \cos( - 90{^\circ}) = 3 \bullet \frac{1}{2} - \sin{60{^\circ}} -\]

\[- \cos{90{^\circ}} =\]

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

\[4)\cos{4455{^\circ}} - \cos( - 945{^\circ}) +\]

\[+ \text{tg}(1035{^\circ}) - ctg( - 1500{^\circ}) =\]

\[= \cos(12 \bullet 360{^\circ} + 90{^\circ} + 45{^\circ}) -\]

\[- \cos( - 3 \bullet 360{^\circ} + 90{^\circ} + 45{^\circ}) +\]

\[+ tg(6 \bullet 180{^\circ} - 45{^\circ}) -\]

\[- \text{ctg}( - 8 \bullet 1440 - 60{^\circ}) =\]

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

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

\[- \text{ctg}( - 60{^\circ}) = - tg\ 45{^\circ} +\]

\[+ ctg\ 60{^\circ} = - 1 + \frac{\sqrt{3}}{3}\]