Question 615316
Since {{{2pi = (12pi)/6}}}, {{{(19pi)/6}}} would be an {{{(7pi)/6}}} more than one full circle. So {{{(19pi)/6}}} and {{{(7pi)/6}}} are co-terminal angles and will have the same sin, cos and tan values.<br>
Since {{{pi = (6pi)/6}}}, {{{(7pi)/6}}} is {{{pi/6}}} more than {{{pi}}}. This makes {{{(7pi)/6}}} (and {{{(19pi)/6}}}) terminate in the 3rd quadrant with a reference angle of {{{pi/6}}}<br>
Since {{{sin(pi/6) = 1/2}}} and since sin is negative in the 3rd quadrant, {{{sin((7pi)/6) = sin((19pi)/6) = -1/2}}}
Since {{{cos(pi/6) = sqrt(3)/2}}} and since cos is negative in the 3rd quadrant, {{{cos((7pi)/6) = cos((19pi)/6) = -sqrt(3)/2}}}
Since {{{tan(pi/6) = (1/2)/(sqrt(3)/2) = 1/sqrt(3) = sqrt(3)/3}}} and since tan is positive in the 3rd quadrant, {{{tan((7pi)/6) = tan((19pi)/6) = sqrt(3)/3}}}