for cos(315)=cos(360-45)=cos(45)=1/sqrt(2) as cos(360-theta)=cos(360)cos(theta)+sin(360)sin(theta)

now cos(360)=1 and sin(360)=0 and theta for above is=45(in degree)

for sin(240)=sin(360-120)=sin(360)cos(120)-cos(360)sin(120)

now using cos(360)=1 and sin(360)=0

we get -sin(120)=-sin(180-60)=-sin(60)=-(sqrt(3)/2)

as sin(180-60)=sin(180)cos(60)-cos(180)sin(60)

use sin(180)=0 and cos(180)=-1

so you get sin(180-60)=sin(60)