Question

Найдите sin a, если cos a = -корень15/4 и 90 градусов < a < 180 градусов

Answer 1

$cos\alpha=-\frac{\sqrt{15}}{4}\\sin\alpha>0\\sin\alpha=\sqrt{1-cos^2\alpha}\\sin\alpha=\sqrt{1-(-\frac{\sqrt{15}}{4})^2}\\sin\alpha=\sqrt{1-\frac{15}{16}}\\sin\alpha=\sqrt{\frac{1}{16}}\\sin\alpha=\frac{1}{4}$

Answer 2

cosa=- √15/4;90°<a<180°

sin²a=1-cos²a=1-15/16=1/16

sina=+√1/16=1/4