Answer
$0$
Work Step by Step
We know that for any acute angle $\theta$, we have: $\sin(\theta)=\cos(90^{o}-\theta)$
Therefore, $\sin(38^{\circ})-\cos(52^{\circ})
\\=\cos(90^{\circ}-38^{ \circ})-\cos(52^{\circ})
\\=\cos 52^{\circ}-\cos \ 52^{\circ}
\\=0$
