Question 1202005
<pre>
The identity to use is {{{cos(x/2)}}}{{{""=""}}}{{{"" +- sqrt((1+cos(x))/2)}}}

Since we are finding {{{cos(x/2)}}} we divide all three sides of the domain for
x by 2, so we'll have the domain for {{{x/2}}}.
 
{{{pi < x < 3pi/2}}}, divide through by 2: 

{{{pi/2 < x/2 < 3pi/4}}}, which in degrees is between 90<sup>o</sup> and 135<sup>o</sup>.

which means x/2 is in QII.  The cosine is negative in QII, so we use the
negative sign

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt((1+cos(x))/2))}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt(((1+("-5/13"))/2))}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt((1-"5/13")/2)}}}

Multiply inside the square root radical by 13/13

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt(expr(((1-"5/13")/2))(13/13))}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt((13-5)/26)}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt(8/26)}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-sqrt(4/13)}}}

{{{cos(x/2)}}}{{{""=""}}}{{{-2/sqrt(13)}}}

You can rationalize the denominator of that if you like.

Edwin</pre>