Найдите: 1 a) sinantg a, если cos a=;2 α=; 2 2 в) cosa и tga, если sin a = √3 2 α= Скачан с vk.com/ma 6) sina utga, если cos a=; α=;3 =- 1 -; г) cosa n tga, если sin a =-.4

Задание 2

В данном задании необходимо найти значения синуса и тангенса угла, зная значение его косинуса или синуса. Используем основные тригонометрические тождества: sin²α + cos²α = 1 и tg α = \(\frac{sin α}{cos α}\).

1. а) cos α = \(\frac{1}{2}\)

   sin²α = 1 - cos²α = 1 - \((\frac{1}{2})^2\) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

   sin α = \(\sqrt{\frac{3}{4}}\) = \(\frac{\sqrt{3}}{2}\)

   tg α = \(\frac{sin α}{cos α}\) = \(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\) = \(\sqrt{3}\)

   Ответ: sin α = \(\frac{\sqrt{3}}{2}\), tg α = \(\sqrt{3}\)

2. б) cos α = \(\frac{1}{3}\)

   sin²α = 1 - cos²α = 1 - \((\frac{1}{3})^2\) = 1 - \(\frac{1}{9}\) = \(\frac{8}{9}\)

   sin α = \(\sqrt{\frac{8}{9}}\) = \(\frac{2\sqrt{2}}{3}\)

   tg α = \(\frac{sin α}{cos α}\) = \(\frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}}\) = 2\(\sqrt{2}\)

   Ответ: sin α = \(\frac{2\sqrt{2}}{3}\), tg α = 2\(\sqrt{2}\)

3. в) sin α = \(\frac{\sqrt{3}}{2}\)

   cos²α = 1 - sin²α = 1 - \((\frac{\sqrt{3}}{2})^2\) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\)

   cos α = \(\sqrt{\frac{1}{4}}\) = \(\frac{1}{2}\)

   tg α = \(\frac{sin α}{cos α}\) = \(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\) = \(\sqrt{3}\)

   Ответ: cos α = \(\frac{1}{2}\), tg α = \(\sqrt{3}\)

4. г) sin α = \(\frac{1}{4}\)

   cos²α = 1 - sin²α = 1 - \((\frac{1}{4})^2\) = 1 - \(\frac{1}{16}\) = \(\frac{15}{16}\)

   cos α = \(\sqrt{\frac{15}{16}}\) = \(\frac{\sqrt{15}}{4}\)

   tg α = \(\frac{sin α}{cos α}\) = \(\frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}}\) = \(\frac{1}{\sqrt{15}}\) = \(\frac{\sqrt{15}}{15}\)

   Ответ: cos α = \(\frac{\sqrt{15}}{4}\), tg α = \(\frac{\sqrt{15}}{15}\)