Question 139726
Can't render theta on this site, so I'll use something else

{{{sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)}}}


So
{{{sin(pi/2+x)=sin(pi/2)cos(x)+cos(pi/2)sin(x)}}}, but {{{sin(pi/2)=1}}} and {{{cos(pi/2)=0}}}, so



{{{sin(pi/2+x)=1*cos(x)+0*sin(x)=cos(x)}}}