If sin(x) = 1/3 and sec(y) = 5/4 , where x and y lie between 0 and pie/2, evaluate the expression cos(x+y).

Sinx=1/3
secy=1/cosy=5/4 => cosy=4/5
cos(x+y)=cosxcosy-sinxsiny =cosx(4/5)-(1/3)siny
so now we need to know cosx and siny i drew two triangles one where sinx=1/3=20/(3*20)
one where cosx=4/5=12*4/(12*5)
 i got the following conclusion: sin(y)=36/60=3/5 and cosx=40(2)^(1/2)/60=2(2)^(1/2)/3
so we have cos(x+y)=2sqrt(2)/3 * 4/5 -1/3 * 3/5 = 8sqrt(2) -3 /15 

The third box is a function

22.