Question 609058
{{{(1/2)cos(pi/12)+(sqrt(3)/2)sin(pi/12)}}}
First of all, when you see/hear "exact value" in a Trig problem, you should know to put your calculator away. The problem can and must be solved using special angles.<br>
But {{{pi/12}}} is NOT one of our special angles!? So how are we supposed to figure this out? Well, somehow we have to find a way to change each {{{pi/12}}} into one of the special angles.<br>
There are a number of Trig properties/identities that allow you to change the argument of a trig function:<ul><li>Angle sum properties: sin(A+B), cos(A+B), tan(A+B)</li><li>Angle difference properties: sin(A-B), cos(A-B), tan(A-B)</li><li>The double angle properties: sin(2x), cos(2x) [of which there are 3 varieties), tan(2x)</li><li>The half angle properties: {{{sin((1/2)x)}}}, {{{cos((1/2)x)}}} and {{{tan((1/2)x)}}}</li></ul>So which one(s) will help us change {{{pi/12}}} into a special angle. There are two clues:<ul><li>None of the properties has a {{{sqrt(3)}}} in it. So your expression, as it is written will not fit any of them. We will need to find a way to rewrite your expression so that it fits the pattern in one of the properties.</li><li>The {{{1/2}}} and the {{{sqrt(3)/2}}} are both special values:
{{{1/2 = sin(pi/6) = cos(pi/3)}}} and
{{{sqrt(3)/2 = sin(pi/3) = cos(pi/6)}}}</li></ul>Let's see what we get if we<ul><li>Replace the 1/2 with {{{sin(pi/6)}}} and the {{{sqrt(3)2}}} with {{{cos(pi/6)}}}:
{{{sin(pi/6)cos(pi/12) + cos(pi/6)sin(pi/12)}}}
This may look like a step backward but upon closer examination we should see that it matches the pattern of the right side sin(A+B):
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
This means we can rewrite
{{{sin(pi/6)cos(pi/12) + cos(pi/6)sin(pi/12)}}}
as
{{{sin(pi/6+pi/12)}}}
This still may not look like progress. But look as what happens when we add the fractions:
{{{sin((2pi)/12+pi/12)}}}
{{{sin((3pi)/12)}}}
{{{sin(pi/4)}}}
And presto! we have a special angle! This simplifies to:
{{{sqrt(2)/2}}}</li><li>Replace the 1/2 with {{{cos(pi/3)}}} and the {{{sqrt(3)2}}} with sin(pi/3):
{{{cos(pi/3)cos(pi/12) + sin(pi/3)sin(pi/12)}}}
This expression marches the pattern of the right side of cos(A-B):
cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
so we can replace 
{{{cos(pi/3)cos(pi/12) + sin(pi/3)sin(pi/12)}}}
with
{{{cos(pi/3 - pi/12)}}}
which simplifies as follows:
{{{cos((4pi)/12 - pi/12)}}}
{{{cos((3pi)/12)}}}
{{{cos(pi/4)}}}
{{{sqrt(2)/2}}}</li></ul>So either way we get {{{sqrt(2)/2}}} for an answer.<br>
NOTE: Trig will be a lot easier if you learn that these properties are patterns. The x's, A's and B's in all these properties are just placeholders. They can be replaced by <i>any mathematical expression</i> and the equation will still be true!!