Question 814362
<pre>
Use a half-angle identity to find the exact value of this expression. 

Given sin{{{(theta)}}} = {{{2sqrt(2)/3}}}, 0° < {{{theta}}} < 90°,

find cos{{{(theta/2)}}} 

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We are given sin{{{(theta)}}} = {{{2sqrt(2)/3}}}

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

{{{2sqrt(2)/3}}} is a positive number, the sine is positive in the first two
 quadrants so 0° < {{{theta}}} < 180° and 0° < {{{theta/2}}} < 90°,

so we take the positive and cos{{{(theta/2)}}} = {{{sqrt((1+cos(theta))/2)}}}

So everything's positive and in the first quadrant.

Since sin{{{(theta)}}} = {{{2sqrt(2)/3}}} and the sine is the opposite over the hypotenuse,
we draw a right triangle containing {{{theta}}} with the length of the opposite side as
the numerator of {{{2sqrt(2)/3}}} and the hypotenuse as the denominator of {{{2sqrt(2)/3}}}. 

Before drawing the right triangle, we calculate the adjacent side using the
Pythagorean theorem:

c² = a² + b²
3² = a² + {{{(2sqrt(2))^2}}}
 9 = a² + 4·2
 9 = a² + 8
 1 = a²
 1 = a 

So the right triangle is like this: 

{{{drawing(850/11,200,-.2,4,-1.5,4.2,rectangle(2.3,0,3,.4),
locate(.6,.5,theta), locate(3.1,2,2sqrt(2)), locate(.77,2.1,3),  
triangle(0,0,3,0,3,4),locate(1.5,0,1)  )}}}

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

From the right triangle, we can get cos{{{(theta)}}} by
using the fact that the cosine is the adjacent over the 
hypotenuse {{{1/3}}}.  Substituting:

cos{{{(theta/2)}}} = {{{sqrt((1+1/3)/2)}}}

To simplify the compound fraction under the radical we multiply
top and bottom by 3

cos{{{(theta/2)}}} = {{{sqrt((3(1+1/3))/3(2))}}}

cos{{{(theta/2)}}} = {{{sqrt((3+1)/6)}}}

cos{{{(theta/2)}}} = {{{sqrt(4/6)}}}

cos{{{(theta/2)}}} = {{{sqrt(2/3)}}}

Rationalize the denominator by multiplying top and bottom by 3

cos{{{(theta/2)}}} = {{{sqrt((3*2)/(3*3))}}}

cos{{{(theta/2)}}} = {{{sqrt(6/9)}}}

cos{{{(theta/2)}}} = {{{sqrt(6)/3)}}}

Edwin</pre>