Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 459

\[\boxed{\mathbf{459.}}\]

\[1)\cos a = \frac{5}{13}\text{\ \ }и\ \ \frac{3\pi}{2} < a < 2\pi\]

\[в\ IV\ четверти:\]

\[\sin a = - \sqrt{1 - \cos^{2}a} =\]

\[= - \sqrt{1 - \left( \frac{5}{13} \right)^{2}} =\]

\[= - \sqrt{\frac{169}{169} - \frac{25}{169}} = - \sqrt{\frac{144}{169}} =\]

\[= - \frac{12}{13}\]

\[tg\ a = \frac{\sin a}{\cos a} = - \frac{12}{13}\ :\frac{5}{13} =\]

\[= - \frac{12}{13} \bullet \frac{13}{5} = - \frac{12}{5}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\left( - \frac{12}{5} \right) =\]

\[= - \frac{5}{12}\]

\[2)\sin a = 0,8\ \ и\ \ \frac{\pi}{2} < a < \pi\]

\[во\ II\ четверти:\]

\[\cos a = - \sqrt{1 - \sin^{2}a} =\]

\[= - \sqrt{1 - {0,8}^{2}} = - \sqrt{1 - 0,64} =\]

\[= - \sqrt{0,36} = - 0,6\]

\[tg\ a = \frac{\sin a}{\cos a} = \frac{0,8}{- 0,6} = - \frac{8}{6} = - \frac{4}{3}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\left( - \frac{4}{3} \right) = - \frac{3}{4}\]

\[3)\ tg\ a = \frac{15}{8}\text{\ \ }и\ \ \pi < a < \frac{3\pi}{2}\]

\[в\ III\ четверти:\]

\[\cos a = - \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}} =\]

\[= - \sqrt{\frac{1}{1 + \left( \frac{15}{8} \right)^{2}}} =\]

\[= - \sqrt{\frac{1}{\frac{64}{64} + \frac{225}{64}}} = - \sqrt{\frac{64}{289}} =\]

\[= - \frac{8}{17}\]

\[\sin a = \cos a \bullet tg\ a = - \frac{8}{17} \bullet \frac{15}{8} =\]

\[= - \frac{15}{17}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\frac{15}{8} = \frac{8}{15}\]

\[4)\ ctg\ a = - 3\ \ и\ \frac{3\pi}{2} < a < 2\pi\]

\[в\ IV\ четверти:\]

\[tg\ a = \frac{1}{\text{ctg\ a}} = \frac{1}{- 3} = - \frac{1}{3}\]

\[\cos a = \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}} =\]

\[= \sqrt{\frac{1}{1 + \left( - \frac{1}{3} \right)^{2}}} = \sqrt{\frac{1}{\frac{9}{9} + \frac{1}{9}}} =\]

\[= \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}\]

\[\sin a = \cos a \bullet tg\ a =\]

\[= \frac{3}{\sqrt{10}} \bullet \left( - \frac{1}{3} \right) = - \frac{1}{\sqrt{10}}\]

\[5)\cos a = 0,8\ \ и\ \ 0 < a < \frac{\pi}{2}\]

\[\ в\ I\ четверти:\]

\[\sin a = \sqrt{1 - \cos^{2}a} =\]

\[= \sqrt{1 - {0,8}^{2}} = \sqrt{1 - 0,64} =\]

\[= \sqrt{0,36} = 0,6\]

\[tg\ a = \frac{\sin a}{\cos a} = \frac{0,6}{0,8} = \frac{6}{8} = \frac{3}{4}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\frac{3}{4} = \frac{4}{3}\]

\[6)\sin a = - \frac{5}{13}\text{\ \ }и\ \ \frac{3\pi}{2} < a < 2\pi\]

\[в\ IV\ четверти:\]

\[\cos a = \sqrt{1 - \sin^{2}a} =\]

\[= \sqrt{1 - \left( - \frac{5}{13} \right)^{2}} =\]

\[= \sqrt{\frac{169}{169} - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}\]

\[tg\ a = \frac{\sin a}{\cos a} = - \frac{5}{13}\ :\frac{12}{13} =\]

\[= - \frac{5}{13} \bullet \frac{13}{12} = - \frac{5}{12}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = 1\ :\left( - \frac{5}{12} \right) =\]

\[= - \frac{12}{5}\]

\[7)\ tg\ a = - 2,4\ \ и\ \ \frac{\pi}{2} < a < \pi\]

\[во\ \text{II\ }четверти:\]

\[tg\ a = - 2,4 = - \frac{24}{10} = - \frac{12}{5}\]

\[\cos a = - \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}} =\]

\[= - \sqrt{\frac{1}{1 + \left( - \frac{12}{5} \right)^{2}}} =\]

\[= - \sqrt{\frac{1}{\frac{25}{25} + \frac{144}{25}}} = - \sqrt{\frac{25}{169}} = - \frac{5}{13}\]

\[\sin a = \cos a \bullet tg\ a =\]

\[= - \frac{5}{13} \bullet \left( - \frac{12}{5} \right) = \frac{12}{13}\]

\[ctg\ a = \frac{1}{\text{tg\ a}} = \frac{1}{- 2,4} = - \frac{10}{24} =\]

\[= - \frac{5}{12}\]

\[8)\ ctg\ a = \frac{7}{24}\text{\ \ }и\ \ \pi < a < \frac{3\pi}{2}\]

\[в\ III\ четверти:\]

\[tg\ a = \frac{1}{\text{ctg\ a}} = 1\ :\frac{7}{24} = \frac{24}{7}\]

\[\cos a = - \sqrt{\frac{1}{1 + tg^{2}\text{\ a}}} =\]

\[= - \sqrt{\frac{1}{1 + \left( \frac{24}{7} \right)^{2}}} =\]

\[= - \sqrt{\frac{1}{\frac{49}{49} + \frac{576}{49}}} = - \sqrt{\frac{49}{625}} =\]

\[= - \frac{7}{25}\]

\[\sin a = \cos a \bullet tg\ a = - \frac{7}{25} \bullet \frac{24}{7} =\]

\[= - \frac{24}{25}\]