Can we use the IVT to conclude that f(x) = sin(x) passes through y = 0 on ?

Yes. The function f is continuous on the closed interval . We have and , therefore and f(b) = 1. Since f(a) = -1 < 0 < 1 = f(b),
the IVT says there is some c in between a and b (that is, ), f(c) = 0.