Question 320808
<pre><b>

If you want {{{cos(theta)/2}}} then since {{{cos(theta)=1/sec(theta)=1/4}}}, then {{{cos(theta)/2}}}{{{"="}}}{{{(1/4)/2=(1/4)*(1/2)=1/8}}}

But I think you want {{{cos(theta/2)}}}.  You need to put parentheses
so we can tell what you want.

We use the half-angle formula:

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

Now {{{cos(theta)}}}{{{"="}}}{{{1/sec(theta)}}}{{{"="}}}{{{1/4}}}

So 

{{{cos(theta/2)}}}{{{"="}}}{{{"" +- sqrt((1+cos(theta))/2)}}}{{{"="}}}{{{"" +- sqrt((1+1/4)/2)}}}{{{"="}}}{{{"" +- sqrt(4*(1+1/4)/(4*2))}}}{{{"="}}}{{{"" +- sqrt((4+1)/8)}}}{{{"="}}}{{{"" +- sqrt(5/8)}}}{{{"="}}}{{{"" +- sqrt((5*2)/(8*2))}}}{{{"="}}}{{{"" +- sqrt(10/16)}}}{{{"="}}}{{{"" +- sqrt(10)/4}}}

Now we can determine the sign by realizing that since {{{theta}}} is in
quadrant I, that half of it, {{{theta/2}}}, will also be in the first quadrant,
and in quadrant I, and all trig functions of angles in quadrant I are positive.

So we have

{{{cos(theta/2)}}}{{{"="}}}{{{sqrt(10)/4}}}

Edwin</pre>