2. Найти sina, tga, sin2a, cos2а, если cos a = - \(\frac{9}{41}\) u \(\frac{\pi}{2}\) <α<π

Ответ: sin α = \(\frac{40}{41}\); tg α = -\(\frac{40}{9}\); sin 2α = -\(\frac{720}{1681}\); cos 2α = -\(\frac{1519}{1681}\)

Так как \(\frac{\pi}{2}\) < α < π, то sin α > 0.

sin² α + cos² α = 1

sin² α = 1 - cos² α = 1 - \(\frac{81}{1681}\) = \(\frac{1600}{1681}\)

sin α = \(\sqrt{\frac{1600}{1681}}\) = \(\frac{40}{41}\)

tg α = \(\frac{sin α}{cos α}\) = \(\frac{40/41}{-9/41}\) = -\(\frac{40}{9}\)

sin 2α = 2 sin α cos α = 2 \(\cdot \frac{40}{41} \cdot\) (- \(\frac{9}{41}\)) = - \(\frac{720}{1681}\)

cos 2α = cos² α - sin² α = \(\frac{81}{1681}\) - \(\frac{1600}{1681}\) = - \(\frac{1519}{1681}\)

Ответ: sin α = \(\frac{40}{41}\); tg α = -\(\frac{40}{9}\); sin 2α = -\(\frac{720}{1681}\); cos 2α = -\(\frac{1519}{1681}\)