1. The problem states that the angle $t$ corresponds to the point $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$ on the unit circle. 2. Recall that on the unit circle, the coordinates of a point are given by $(\cos(t), \sin(t))$ where $t$ is the angle measured from the positive x-axis. 3. Therefore, for the point $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$, we have: $$\cos(t) = -\frac{\sqrt{2}}{2}$$ $$\sin(t) = -\frac{\sqrt{2}}{2}$$ 4. This means the cosine and sine of the angle $t$ are both $-\frac{\sqrt{2}}{2}$. 5. The point lies in the third quadrant where both sine and cosine are negative, which matches our values. Final answer: $$\cos(t) = -\frac{\sqrt{2}}{2}, \quad \sin(t) = -\frac{\sqrt{2}}{2}$$