![\bf sin^2(\theta )=\cfrac{3}{4}\implies [sin(\theta )]^2=\cfrac{3}{4}\implies \sqrt{[sin(\theta )]^2}=\sqrt{\cfrac{3}{4}}\implies sin(\theta )=\cfrac{\sqrt{3}}{\sqrt{4}} \\\\\\ sin(\theta )=\cfrac{\sqrt{3}}{2}\qquad \qquad \stackrel{\textit{and if you check on your handy dandy \underline{Unit Circle}}}{\theta =\cfrac{\pi }{3}~~~~\textit{on I Quadrant},\qquad \theta =\cfrac{2\pi }{3}~~~~\textit{on II Quadrant}}](/tpl/images/0335/4688/66aa2.png)
Idon’t remember learning this so would be greatly appreciated
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Mathematics, 19.10.2019 22:30 griffislandon74
Idon’t remember learning this so would be greatly appreciated
Answers: 2
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