Question 1160204
.


            Regarding part  (a),  the solution and the answer by @Mathlover1 are both incorrect;


            therefore, I came to bring the correct solution.



<pre>
(a)  if  {{{tan(theta)}}} = -{{{5/12}}}   and  {{{theta}}} is in QVI, then'


     {{{cos(theta)}}} = {{{12/13}}};


     {{{theta/2}}}  lies in QII, where sine is positive,


     and therefore  {{{sin(theta/2)}}} = + {{{sqrt((1-cos(theta))/2)}}} = {{{sqrt((1-12/13)/2)}}} = {{{sqrt(1/26)}}} = {{{sqrt(26)/26}}},


      totally different formula than solution by @MathLover1.
</pre>


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