Question 1201608
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If sine theta = 1/3, theta in Quadrant II, find sin theta + pi/6
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<pre>
Step 1.  If {{{sin(theta)}}} = 1/3  and  {{{theta}}} in Quadrant II, then it implies

            {{{cos(theta)}}} = {{{-sqrt(1-sin^2(theta))}}} = {{{-sqrt(1 - 1/9)}}} = {{{-sqrt(8)/3}}} = {{{(-2sqrt(2))/3}}}.


Step 2.  {{{sin(theta+pi/6)}}} = {{{sin(theta)*cos(pi/6)}}} + {{{cos(theta)*sin(pi/6)}}} = 

         = {{{(1/3)*(sqrt(3)/2)}}} + {{{((-2*sqrt(2))/3)*(1/2)}}} = {{{sqrt(3)/6}}} - {{{(2*sqrt(2))/6}}} = {{{(sqrt(3)-2*sqrt(2))/6}}} = -0.1827 (rounded).    <U>ANSWER</U>
</pre>

Solved.