Come provare che #sin (pi / 2 - theta) # = #cos theta #?

using appropriate #color(blue)" Addition formula " #

#• sin(A ± B) = sinAcosB ± cosAsinB #

hence # sin(pi/2 -theta) = sin(pi/2) costheta - cos(pi/2)sintheta #

now # sin(pi/2) = 1 " and " cos(pi/2) = 0 #

hence # sin(pi/2)costheta - cos(pi/2)sintheta = costheta - 0 #

#rArr sin(pi/2 - theta ) = costheta #