346. В треугольнике ABC угол B - прямой. Найдите sin A, если: a) cos A = \frac{\sqrt{3}}{2}; б) cos A = 0,6; в) cos A = \frac{3\sqrt{39}}{20}; г) cos A = \frac{\sqrt{21}}{5}

Решение: Используем основное тригонометрическое тождество: \(sin^2 A + cos^2 A = 1\). a) Если \(cos A = \frac{\sqrt{3}}{2}\), то: \[sin^2 A = 1 - cos^2 A = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4}\] \[sin A = \sqrt{\frac{1}{4}} = \frac{1}{2}\] б) Если \(cos A = 0,6 = \frac{3}{5}\), то: \[sin^2 A = 1 - cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}\] \[sin A = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0,8\] в) Если \(cos A = \frac{3\sqrt{39}}{20}\), то: \[sin^2 A = 1 - cos^2 A = 1 - \left(\frac{3\sqrt{39}}{20}\right)^2 = 1 - \frac{9 \cdot 39}{400} = 1 - \frac{351}{400} = \frac{49}{400}\] \[sin A = \sqrt{\frac{49}{400}} = \frac{7}{20}\] г) Если \(cos A = \frac{\sqrt{21}}{5}\), то: \[sin^2 A = 1 - cos^2 A = 1 - \left(\frac{\sqrt{21}}{5}\right)^2 = 1 - \frac{21}{25} = \frac{4}{25}\] \[sin A = \sqrt{\frac{4}{25}} = \frac{2}{5} = 0,4\] Ответы: a) \(sin A = \frac{1}{2}\) б) \(sin A = 0,8\) в) \(sin A = \frac{7}{20}\) г) \(sin A = \frac{2}{5} = 0,4\)