Question 868352
Since u and v are between 0 and pi/2, this means that they're both in quadrant I


If sin(u) = 2/3, then....


sin^2(u) + cos^2(u) = 1
(2/3)^2 + cos^2(u) = 1
4/9 + cos^2(u) = 1
cos^2(u) = 1 - 4/9
cos^2(u) = 9/9 - 4/9
cos^2(u) = 5/9
cos(u) = sqrt(5/9)  ... ignore the negative square root since cos(u) > 0 in quadrant I
cos(u) = sqrt(5)/sqrt(9)
cos(u) = sqrt(5)/3 


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If sin(v) = 1/3, then....


sin^2(v) + cos^2(v) = 1
(1/3)^2 + cos^2(v) = 1
1/9 + cos^2(v) = 1
cos^2(v) = 1 - 1/9
cos^2(v) = 9/9 - 1/9
cos^2(v) = 8/9
cos(v) = sqrt(8/9)  ... ignore the negative square root since cos(v) > 0 in quadrant I
cos(v) = sqrt(8)/sqrt(9)
cos(v) = sqrt(8)/3 
cos(v) = sqrt(4*2)/3 
cos(v) = sqrt(4)*sqrt(2)/3 
cos(v) = 2*sqrt(2)/3 
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From those two parts above, we have found cos(u) = sqrt(5)/3 and cos(v) = 2*sqrt(2)/3 


So all together, we know this:


sin(u) = 2/3
sin(v) = 1/3
cos(u) = sqrt(5)/3
cos(v) = 2*sqrt(2)/3 


We use them to plug them into the formula below (and simplify)


sin(u+v) = sin(u)*cos(v) + cos(u)*sin(v)
sin(u+v) = (2/3)*(2*sqrt(2)/3) + (sqrt(5)/3)*(1/3)
sin(u+v) = 4*sqrt(2)/9 + sqrt(5)/9
sin(u+v) = (4*sqrt(2) + sqrt(5))/9


So the final answer is {{{sin(u+v) = (4*sqrt(2) + sqrt(5))/9}}}