Question

$9 \sqrt{5} \sin(2x)$
Если синус х =-2/3 и -п<2х<п​

Answer 1

$-\pi <2x<\pi \Rightarrow -\frac{\pi }{2}<x<\frac{\pi }{2}\\\\Cosx>0\\\\Sinx=-\frac{2}{3}\\\\Cosx=\sqrt{1-Sin^{2}x }=\sqrt{1-(-\frac{2}{3})^{2}}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9} }=\frac{\sqrt{5}}{3}\\\\9\sqrt{5}Sin2x=9\sqrt{5}*2Sinx Cosx= 18\sqrt{5}*(-\frac{2}{3})*\frac{\sqrt{5}}{3} =-20\\\\Otvet:\boxed{-20}$