Answer
$sin(x+y)=\frac{1}{15}(4+6\sqrt 2)$
Work Step by Step
Evaluate the expression $sin(x+y)$
Given: $sinx=\frac{1}{3}$ and $secy=\frac{5}{4}$
$sin(x+y)=sinxcosy+cosxsiny$ ...(1)
Thus,
$sinx=\frac{1}{3}$ gives opp =1 , hyp = 3 and adj $=\sqrt {3^{2}-1^{2}}=2\sqrt 2$
Therefore, $cos x=\frac{2\sqrt 2}{3}$
Now, $secy=\frac{5}{4}$ gives hyp =5 , adj =4
and opp $=\sqrt {5^{2}-3^{2}}=3$
Therefore, $siny=\frac{3}{5}$ and $cosy=\frac{4}{5}$
Equation (1) becomes
$sin(x+y)=(\frac{1}{3})(\frac{4}{5})+(\frac{2\sqrt 2}{3})(\frac{3}{5})$
$=\frac{4}{15}+\frac{6\sqrt 2}{15}$
Hence, $sin(x+y)=\frac{1}{15}(4+6\sqrt 2)$