Question 1026266
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Assume that a is an angle in standard position whose terminal side contains the point (sqrt(6), -sqrt(2)). 
Find the exact values of the 6 trig. functions
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<pre>
The radius-vector is  r^2 = 6 + 2 = 8;  r  {{{sqrt(8)}}} = {{{2*sqrt(2)}}}.


{{{sin(theta)}}} = {{{y/r}}} = {{{-sqrt(2)/(2*sqrt(2))}}} = {{{-1/2}}}.


{{{cos(theta)}}} = {{{x/r}}} = {{{sqrt(6)/(2*sqrt(2))}}} = {{{sqrt(3)/2}}}.


{{{tan(theta)}}} = {{{sin(theta)/cos(theta)}}} = {{{-1/sqrt(3)}}} = {{{-sqrt(3)/3}}}.


The angle  {{{theta}}}  is  {{{-pi/6}}} = -30 degrees.   Or the same as  {{{theta}}} = {{{11pi/6}}} = 330 degrees.
</pre>

Solved.