If sin^2 of theta = 1/2 and pi< theta < 3pi/2 (quadrant III) then theta= ? I have no idea how to do these type of problems! Please Help!
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$\displaystyle \sin^2 \theta = \frac {1} {2} \Rightarrow \sin \theta= \frac {1}{\sqrt{2}} \Rightarrow \theta= -\frac {\pi} {4} \Rightarrow \theta= \pi + \frac{\pi}{4} \Rightarrow \theta= 5\frac{\pi}{4} $
Originally Posted by askmemath $\displaystyle \sin^2 \theta = \frac {1} {2} \Rightarrow \sin \theta= \frac {1}{\sqrt{2}} \Rightarrow \theta= -\frac {\pi} {4} \Rightarrow \theta= \pi + \frac{\pi}{4} \Rightarrow \theta= 5\frac{\pi}{4} $ sine of theta must be taken as negative sqaure root of half here. then the value of theta will be 5Pi/4
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