Question

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

Answer 1

`\sin 2 \theta = -(4\sqrt 5)/(9)`

Explanation:

`\text{Given that,}\\\\~~~~~~\sin \theta = \frac 23\\\\\implies \sin^2 \theta = \left( \frac 23 \right)^2\\\\\implies 1- \cos^2 \theta = \frac 49 \\\\\implies \cos^2 \theta = 1 - \frac 49 \\\\\implies \cos^2 \theta = \frac 59\\\\\implies \cos \theta = \pm√(\frac 59) \\\\\implies \cos \theta = \pm \frac{\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 = - (\sqrt 5 )/(3)\\\\\text{Now,}\\\\\sin 2\theta = 2\sin \theta \cos \theta \\\\\\~~~~~~~~=2 \cdot \frac 23 \cdot \left( -(\sqrt 5)/(3) \right)\\\\\\~~~~~~~~=-(4\sqrt 5)/(9)`

`\textbf{Quadrant rule:}\\\\`