Precalculus, Chapter 4, 4.2, Section 4.2, Problem 41

Given
sin(t) = 4/5


a) sin(pi - t)
this is of the form sin(a-b) where a= pi and b= t
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
so,
sin(pi - t)=sin(pi)cos(t) - cos(pi)sin(t) = 0 - (-1)sin(t) = sin(t)
or simply we know by the identity sin(pi-t) = sin(t) we get

sin(pi-t) = sin(t) = 4/5

b) sin(t + pi)
similar to above we get
sin(t + pi)= sin(pi+t) = -sin(t)
we can also get this by the identity
so now ,
sin(t + pi) = - sin(t) = -(4/5)