Question 1036142
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Find sin 2 theta, cos 2 theta, and tan 2 theta for each set of conditions.

cos theta = -5/13 and pi/2 < theta < pi.
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Since  {{{cos(theta)}}} = {{{-5/13}}}  and  {{{pi/2}}} < {{{theta}}} < {{{pi}}}, 

{{{sin(theta)}}} = {{{sqrt(1 - (-5/13)^2)}}} = {{{sqrt(1 - 25/169)}}} = {{{sqrt((169-25)/169)}}} = {{{sqrt(144/169)}}} = {{{12/13}}}.

The sign "+" was chosen at the square root because the sine is positive in the second quadrant.


Next,

{{{sin(2*theta))}}} = {{{2*sin(theta)*cos(theta)}}} = {{{2*(12/13)*(-5/13))}}} = {{{-120/169)}}},


{{{cos(2*theta))}}} = {{{cos^2(theta) - sin^2(theta)}}} = {{{(-5/13)^2 - (12/13)^2}}} = {{{25/169 - 144/169}}} = {{{-119/169}}}.


By the way, the signs of the  {{{sin(2theta)}}}  and  {{{cos(2theta)}}}  say  that {{{2theta}}}  lies in the quadrant 3.


Now,  {{{tan(2*theta)}}} = {{{sin(2*theta)/cos(2*theta)}}} = {{{((-120/169))/((-119/169))}}} = {{{120/119}}}.

Solved.
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