If sinθ + cosθ = 1,

Given $\sin \theta+\cos \theta=1$

By squaring both sides, we get

$(\sin \theta+\cos \theta)^{2}=(1)^{2}$

i. e. $\sin 2 \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=1$

i. e. $1+2 \sin \theta \cos \theta=1$

i.e. $2 \sin \theta \cos \theta=0$

i.e. $\sin 2 \theta=0$

Hence, value of $\sin 2 \theta$ is 0 .