Geometry -- 2

$\begin{aligned} \cos \left(\frac \pi 2 + 2\theta \right) & = 1- 2\sin^2 \left(\frac \pi 4 + \theta \right) \\ & = 1 - \frac 29 = \frac 79 & \small \color{#3D99F6} \text{Since }\cos \left(\frac \pi 2 - x\right) = \sin x \\ \sin (-2\theta) & = \frac 79 & \small \color{#3D99F6} \text{and }\sin (-x) = - \sin x \\ - \sin (2\theta) & = \frac 79 \\ \implies \sin (2\theta) & = \boxed{-\frac 79} \end{aligned}$

$\sin {2\theta} = -\cos {(\frac {\pi}{2} + \theta)}$ $\quad$ [Since $\sin {(\frac {\pi}{4} + 2\theta)} = \frac 13$ which is smaller than $\sin {\frac {\pi}{4}}$ meaning the vector is in the Second Quadrant and has a negative cosine value] $\\$ $\\$

$-\cos {(\frac {\pi}{2} + \theta)} = -\cos {[2 (\frac {\pi}{4} + \theta]} = 2 \sin ^2 {\frac {\pi}{4} + \theta} - 1 = \frac 29 - 1 = - \frac 79$ $\\$ $\\$