Задание 1. Найдите: 1 a) cosa и tga, если sina = 5; 3 б) sinß и tgß, если соѕẞ== 7

Ответ: а) cosα = \(\frac{2\sqrt{6}}{5}\), tgα = \(\frac{\sqrt{6}}{12}\); б) sinβ = \(\frac{2\sqrt{10}}{7}\), tgβ = \(\frac{2\sqrt{10}}{3}\)

Решение:

а) Дано: sinα = \(\frac{1}{5}\). Найти: cosα, tgα.

• Основное тригонометрическое тождество: sin²α + cos²α = 1
• cos²α = 1 - sin²α = 1 - \((\frac{1}{5})^2\) = 1 - \(\frac{1}{25}\) = \(\frac{24}{25}\)
• cosα = ±\(\sqrt{\frac{24}{25}}\) = ±\(\frac{2\sqrt{6}}{5}\). Так как угол острый, cosα > 0.
• cosα = \(\frac{2\sqrt{6}}{5}\)
• tgα = \(\frac{sinα}{cosα}\) = \(\frac{1/5}{2\sqrt{6}/5}\) = \(\frac{1}{2\sqrt{6}}\) = \(\frac{\sqrt{6}}{12}\)

б) Дано: cosβ = \(\frac{3}{7}\). Найти: sinβ, tgβ.

• sin²β = 1 - cos²β = 1 - \((\frac{3}{7})^2\) = 1 - \(\frac{9}{49}\) = \(\frac{40}{49}\)
• sinβ = ±\(\sqrt{\frac{40}{49}}\) = ±\(\frac{2\sqrt{10}}{7}\). Так как угол острый, sinβ > 0.
• sinβ = \(\frac{2\sqrt{10}}{7}\)
• tgβ = \(\frac{sinβ}{cosβ}\) = \(\frac{2\sqrt{10}/7}{3/7}\) = \(\frac{2\sqrt{10}}{3}\)

Ответ: а) cosα = \(\frac{2\sqrt{6}}{5}\), tgα = \(\frac{\sqrt{6}}{12}\); б) sinβ = \(\frac{2\sqrt{10}}{7}\), tgβ = \(\frac{2\sqrt{10}}{3}\)