We use the identities:
$$
\tag{1}\cos u =\sin\Bigl({\pi\over2}-u\Bigr)
$$
and$$
\tag {2}\sin u +\sin v= { 2}\sin\Bigl({u+v\over 2} \Bigr)\cos\Bigl({u-v\over 2} \Bigr).
$$
We have, by (1): $$
\sin x+\cos y= \color{maroon}{ \sin x +\sin \Bigl( {\pi\over2}-y\Bigr)}.$$
From (2):$$
\eqalign{
\color{maroon}{
\sin x +\sin\Bigl ( {\pi\over2}-y\Bigr)}&=
{ 2}\sin\Bigl({{\pi\over2}+x-y\over 2} \Bigr)\cos\Bigl( {x-{\pi\over2}+ y\over 2}\Bigr )\cr
&=\color{darkgreen}{{ 2}\sin\Bigl(\textstyle{{\pi\over4}+{x\over2}-{y\over 2}} \Bigr)
\cos\Bigl(\textstyle {{x\over2}-{\pi\over4}+ {y\over 2}}\Bigr ) }
.
}
$$
Using (1) again:
$$\color{darkgreen}{
{ 2}\sin\Bigl(\textstyle{{\pi\over4}+{x\over2}-{y\over 2}} \Bigr)
\cos\Bigl(\textstyle {{x\over2}-{\pi\over4}+ {y\over 2}}\Bigr ) }
= { 2}\sin\Bigl( \textstyle{{\pi\over4}+{x\over2}-{y\over 2}} \Bigr)
\color{orange}{\sin\Bigl( { {\pi\over4 }-{x\over2}- {y\over 2}}\Bigr )}.
$$
Since $\sin({\pi\over2}+\theta)=\sin({\pi\over2}-\theta)$:
$$
\color{orange}{\sin\Bigl( \textstyle{ {\pi\over4 }-{x\over2}- {y\over 2}}\Bigr ) }
=\sin\Bigl( \textstyle{ {\pi\over4 }+{x\over2}+ {y\over 2}}\Bigr ) .
$$
Putting everything together gives your result.