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Find the point (x, y) on the unit circle that corresponds to the real number 5pi/6. Use it to evaluate cos t. cos t = (-sqrt(3)/2)?
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by definition, we can write: \[\Large \begin{gathered} x = \cos \left( {\frac{{5\pi }}{6}} \right) = ...? \hfill \\ \hfill \\ y = \sin \left( {\frac{{5\pi }}{6}} \right) = \sqrt {1 - {{\left\{ {\cos \left( {\frac{{5\pi }}{6}} \right)} \right\}}^2}} = ...? \hfill \\ \end{gathered} \]
for the y-coordinate I choose the positive square rooth since your point is at II quadrant
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