Question 992753
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Given that cotx=-5/6 and x is in quadrant 2, find sin2x, cos2x,tan2x
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cot(x) = {{{-5/6}}}     --->     {{{cos(x)/sin(x)}}} = {{{-5/6}}} ---> {{{cos(x)}}} = {{{-5/6}}}.{{{sin(x)}}}     ---> 


{{{cos^2(x)}}} = {{{25/36}}}.{{{sin^2(x)}}}.         (1)


{{{cos^2(x)}}} + {{{sin^2(x)}}} = 1,                 (2)


Now substitute  (1)  into  (2).  You will get


{{{(1 + (25/36))}}}.{{{sin^2(x)}}} = 1,     or


{{{sin^2(x)}}} = {{{36/(25+36)}}} = {{{36/61}}}.


Hence,   sin(x) = {{{6/sqrt(61)}}},   cos(x) = -{{{5/sqrt(61)}}}     (taking into account that  x  is in the second quadrant).


Now,  when you know  sin(x)  and  cos(x)  values,  you can calculate the remaining functions.


Use the trigonometry formulas


sin(2x) = 2sin(x)*cos(x), = . . . ,


cos(2x) = {{{cos^2(x)}}} - {{{sin^2(x)}}} = . . . ,


and then  tan(2x) = {{{sin(2x)/cos(2x)}}} = . . . .


Complete these calculations yourself.


Good luck!