Question 1113736
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1.  If csc(a) = 3 and the angle "a" is in QI, it implies that


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



2.  If tan(b) = -7  and the angle "b" is in QII, it implies that


    cos^2(b) = {{{1/(1+tan^2(b))}}} = {{{1/(1+(-7)^2)}}} = {{{1/50}}}  and  cos(b) = -{{{sqrt(1/50)}}} = {{{-sqrt(50)/50}}} = {{{-(5*sqrt(2))/50}}} = {{{-sqrt(2)/10}}}.

    Notice that cos(b) is negative, since "b" is in QII.


    Next, knowing cos(b), you can calculate


    sin(b) = {{{sqrt(1-cos^2(b))}}} = {{{sqrt(1-2/100)}}} = {{{sqrt(98/100)}}} = {{{(7*sqrt(2))/10}}}.


<U>Answer</U>.  sin(a) = {{{1/3}}}.  cos(a) = {{{(2*sqrt(2))/3}}}.  sin(b) = {{{(7*sqrt(2))/10}}}.  cos(b) = {{{-sqrt(2)/10}}}.
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