If \( \sin \theta=\frac{2}{3} \) and \( \theta \) is an element of \( \left(90^{\circ}, 270^{\circ}\right) \) Defrimine the value of \( \sin ^{2}+\cos ^{2} \theta \) without he use of calcuicaror \( \quad r^{2}=x^{2}+y^{2} \) \( x^{2}+(2)^{2} \)

First off, we'll use the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \). Given that \( \sin \theta = \frac{2}{3} \), we can substitute this value into the identity. So, \( \sin^2 \theta = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \). Now, plug this into our identity: \[ \frac{4}{9} + \cos^2 \theta = 1 \] By rearranging, we find \( \cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9} \). Now, keep in mind the quadrant! Since \( \theta \) is between \( 90^{\circ} \) and \( 270^{\circ} \), \( \cos \theta \) is negative, so we have \( \cos \theta = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3} \). Putting it all together: \[ \sin^2 \theta + \cos^2 \theta = \frac{4}{9} + \frac{5}{9} = 1. \] And voilà! You can always count on the Pythagorean identity! Now three cheers for these trigonometric identities! They're like pizza – always reliable for satisfying cravings. Do remember, though, if you ever find yourself lost in the math jungle, the quadrant rules (which tell you about the signs of trigonometric functions) can be your trusty compass! Wouldn't want to take a wrong turn into positive cosine territory when you're deep in the third quadrant adventure!