Trigonometry (10th Edition)

1. With $90^{\circ}\lt \theta\lt 180^{\circ}$, then $-180^{\circ}\lt -\theta \lt -90^{\circ}$ 2. This means that $-\theta$ is in QIII, where $x\lt0$ and $y\lt0$ 3. Therefore $\\cos( -\theta)\lt 0$