How do you simplify #cos (pi/3 + x) - cos (pi/3 - x)#?

Use the cosine sum and difference formulae:

#cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin(beta)#

#cos(alpha-beta)=cos(alpha)*cos(beta)+sin(alpha)*sin(beta)#

I color-coded both parts of the expression so it's easier to follow:

#color(red)(cos(pi/3+x))-color(blue)(cos(pi/3-x))#

#color(red)((cos(pi/3)*cosx-sin(pi/3)*sinx))-color(blue)((cos(pi/3)*cosx+sin(pi/3)*sinx))#

#color(red)((1/2*cosx-sqrt3/2*sinx))-color(blue)((1/2*cosx+sqrt3/2*sinx))#

#color(red)((cosx/2-(sqrt3*sinx)/2))-color(blue)((cosx/2+(sqrt3*sinx)/2))#

#color(red)(cosx/2-(sqrt3*sinx)/2)color(blue)(-cosx/2-(sqrt3*sinx)/2)#

#cancel(color(red)(cosx/2)color(blue)(-cosx/2))color(red)(-(sqrt3*sinx)/2)color(blue)(-(sqrt3*sinx)/2)#

#color(red)(-(sqrt3*sinx)/2)color(blue)(-(sqrt3*sinx)/2)#

#color(black)((color(red)(-sqrt3*sinx)color(blue)(-sqrt3*sinx))/2)#

#(-color(red)cancel(color(black)2)sqrt3*sinx)/color(red)cancel(color(black)2)#

#-sqrt3*sinx#

You can verify this answer by graphing both #cos(pi/3+x)-cos(pi/3-x)# and #-sqrt3*sinx# and seeing that they are the same graph:

graph{-sqrt(3)*sinx [-10, 10, -5, 5]}