Question 1042080
Use the Pythagorean Trig Identity to solve for {{{sin(theta)}}}



{{{sin^2(theta) + cos^2(theta) = 1}}}



{{{sin^2(theta) + (cos(theta))^2 = 1}}}



{{{sin^2(theta) + (-1/3)^2 = 1}}}



{{{sin^2(theta) + 1/9 = 1}}}



{{{sin^2(theta) + 1/9-1/9 = 1-1/9}}}



{{{sin^2(theta) = 9/9-1/9}}}



{{{sin^2(theta) = 8/9}}}



{{{sqrt(sin^2(theta)) = sqrt(8/9)}}}



{{{abs(sin(theta)) = sqrt(8/9)}}}



{{{sin(theta) = ""+-sqrt(8/9)}}}



{{{sin(theta) = -sqrt(8/9)}}} See note below



{{{sin(theta) = -(sqrt(8))/(sqrt(9))}}}



{{{sin(theta) = -(sqrt(8))/(3)}}}



{{{sin(theta) = -(sqrt(4*2))/(3)}}}



{{{sin(theta) = -(sqrt(4)*sqrt(2))/(3)}}}



{{{sin(theta) = -(2*sqrt(2))/(3)}}}



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Note: 



θ is in the interval [π, 3π/2], so theta is in quadrant 3 where sine is negative. 



So {{{sin(theta)<0}}} which means we can drop the plus/minus and focus solely on the minus. 



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The final answer is {{{sin(theta) = -(2*sqrt(2))/(3)}}}