Question

Найти sin2a, если 90°

Answer 1

Ответ:

sin2a = 2sin(a)cos(a)

sin a = 5/13

$cosa = \sqrt{1 - sin^{2}a } = \sqrt{1-\frac{25}{169} } =\sqrt{\frac{144}{169} } =-\frac{12}{13}$

$2sin(a)cos(a)= 2 * \frac{5}{13} *(-\frac{12}{13}) = -\frac{120}{169}$

Answer 2

α - угол второй четверти , значит Cosα < 0 .

$Cos\alpha=-\sqrt{1-Sin^{2}\alpha} =-\sqrt{1-(\frac{5}{13})^{2}} =-\sqrt{1-\frac{25}{169}} =-\sqrt{\frac{144}{169}}=\boxed{-\frac{12}{13}}\\\\\\Sin2\alpha=2Sin\alpha Cos\alpha=2*\frac{5}{13}*(-\frac{12}{13})=-\frac{120}{169} \\\\\boxed{Sin2\alpha=-\frac{120}{169}}$