Question

Find sin20 if sin 0 = 3; 90° < 0 < 180°

Answer 1

Answer:

$\sin 2 \theta = -\dfrac{4\sqrt 5}{9}$

Step-by-step explanation:

$\text{Given that,}\\\\~~~~~~\sin \theta = \dfrac 23\\\\\implies \sin^2 \theta = \left( \dfrac 23 \right)^2\\\\\implies 1- \cos^2 \theta = \dfrac 49 \\\\\implies \cos^2 \theta = 1 - \dfrac 49 \\\\\implies \cos^2 \theta = \dfrac 59\\\\\implies \cos \theta = \pm\sqrt{\dfrac 59} \\\\\implies \cos \theta = \pm \dfrac{\sqrt 5 }3\\\\\text{Since}~ 90^\circ < \theta < 180^{\circ},~\text{the angle lies in 2nd quadrant}\left(\cos \theta ~\text{is negative}\right).$

$\text{So,}~ \cos \theta = - \dfrac{\sqrt 5 }{3}\\\\\text{Now,}\\\\\sin 2\theta = 2\sin \theta \cos \theta \\\\\\~~~~~~~~=2 \cdot \dfrac 23 \cdot \left( -\dfrac{\sqrt 5}{3} \right)\\\\\\~~~~~~~~=-\dfrac{4\sqrt 5}{9}$

$\textbf{Quadrant rule:}$