Question 1011267

{{{sin((pi/2 - x) + y) = cos(x-y)}}}

start with left side and arrive to right side:


{{{sin((pi/2 - x)) sin(y)+cos((pi/2 - x)) cos(y)}}}


since 

{{{sin(pi/2-x) = cos(x) sin(pi/2)-cos(pi/2) sin(x)}}}
{{{cos(pi/2-x) = cos(pi/2) cos(-x)-sin(pi/2) sin(-x)}}}


we have


={{{(cos(x) sin(pi/2)-cos(pi/2) sin(x)) sin(y)+(cos(pi/2) cos(-x)-sin(pi/2) sin(-x)) cos(y)}}}

={{{cos(x) sin(y)sin(pi/2)-cos(pi/2) sin(y)sin(x) +cos(pi/2)cos(y) cos(-x)-sin(pi/2)cos(y) sin(-x)}}} ...since {{{sin(pi/2)=1}}} and {{{cos(pi/2)=0}}}, we have


={{{cos(x) sin(y)*1-0* sin(y)sin(x) +0*cos(y) cos(x)-1*cos(y) sin(-x)}}} 


={{{cos(x) sin(y) -cos(y) sin(-x) =cos(x-y)}}}...since {{{sin(-x) =-sin(x)}}}, we have 


={{{cos(x)sin(y)-(-sin(x) cos(y))}}}

={{{cos(x)sin(y)-sin(x) cos(y)}}}

={{{cos(x-y)}}}