1. **Stating the problem:** We have two functions, $y=\cos x$ and $y=\sin x$, graphed on the Cartesian plane. The curves intersect at points that create three shaded regions labeled $R_1$, $R_2$, and $R_3$. We want to understand these regions formed by the intersections of $\cos x$ and $\sin x$. 2. **Finding intersection points:** To find where the curves intersect, set $\cos x = \sin x$. 3. **Solve the equation:** $$\cos x = \sin x$$ Divide both sides by $\cos x$ (assuming $\cos x \neq 0$): $$\cancel{\cos x} \cdot \frac{1}{\cancel{\cos x}} = \sin x \cdot \frac{1}{\cos x} \implies 1 = \tan x$$ 4. **Find solutions for $\tan x = 1$:** $$x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}$$ The two closest intersection points around zero are at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. 5. **Analyze regions:** - $R_1$ is between the curves near the top-left quadrant, between $x = -\frac{3\pi}{4}$ and $x = \frac{\pi}{4}$. - $R_2$ is between $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. - $R_3$ is below the x-axis in the bottom-right quadrant, between $x = \frac{5\pi}{4}$ and $x = \frac{9\pi}{4}$. 6. **Summary:** The curves $y=\cos x$ and $y=\sin x$ intersect at $x=\frac{\pi}{4} + k\pi$, creating alternating regions $R_1$, $R_2$, and $R_3$ between these points. Each region is bounded by the two curves between consecutive intersection points.