1. **State the problem:** We are given two sinusoidal functions $y=f(x)$ and $y=g(x)$ graphed on a Cartesian coordinate system with $x$ in degrees and $y$ values ranging approximately from -3 to 3. 2. **Analyze $f(x)$:** The function $f(x)$ starts at $(0,0)$, reaches a maximum near $(90^\circ, 2)$, crosses zero at $180^\circ$, reaches a minimum near $(270^\circ, -3)$, and returns to zero at $360^\circ$. This suggests a sinusoidal function with period $360^\circ$, amplitude about 3, and vertical shift 0. 3. **General form for $f(x)$:** Since it starts at zero and reaches max at $90^\circ$, $f(x)$ resembles a sine function with amplitude $A$, period $360^\circ$, and vertical shift $D=0$: $$f(x) = A \sin\left(\frac{2\pi}{360} x\right) = A \sin\left(\frac{\pi}{180} x\right)$$ 4. **Find amplitude $A$ for $f(x)$:** The max near $90^\circ$ is about 2, min near $270^\circ$ is about -3. The amplitude is half the distance between max and min: $$A = \frac{2 - (-3)}{2} = \frac{5}{2} = 2.5$$ 5. **Adjust for vertical shift:** The midline is average of max and min: $$D = \frac{2 + (-3)}{2} = \frac{-1}{2} = -0.5$$ 6. **Rewrite $f(x)$ with amplitude and vertical shift:** $$f(x) = 2.5 \sin\left(\frac{\pi}{180} x\right) - 0.5$$ 7. **Analyze $g(x)$:** The function $g(x)$ starts at $(0,0)$, reaches max near $(90^\circ, 3)$, crosses $f(x)$ at $A$ near $(90^\circ, 2)$, crosses zero at $180^\circ$, reaches min near $(270^\circ, -2)$, crosses $f(x)$ again at $B$ near $(270^\circ, -2)$, and returns to zero at $360^\circ$. 8. **General form for $g(x)$:** Since $g(x)$ also has period $360^\circ$ and zero at $0^\circ$, $180^\circ$, and $360^\circ$, it resembles a sine function with amplitude $A_g$ and vertical shift $D_g=0$: $$g(x) = A_g \sin\left(\frac{\pi}{180} x\right)$$ 9. **Find amplitude $A_g$ for $g(x)$:** Max near 3, min near -2, so amplitude: $$A_g = \frac{3 - (-2)}{2} = \frac{5}{2} = 2.5$$ 10. **Check vertical shift for $g(x)$:** Midline: $$D_g = \frac{3 + (-2)}{2} = \frac{1}{2} = 0.5$$ 11. **Rewrite $g(x)$ with amplitude and vertical shift:** $$g(x) = 2.5 \sin\left(\frac{\pi}{180} x\right) + 0.5$$ 12. **Final expressions:** $$\boxed{f(x) = 2.5 \sin\left(\frac{\pi}{180} x\right) - 0.5}$$ $$\boxed{g(x) = 2.5 \sin\left(\frac{\pi}{180} x\right) + 0.5}$$ These functions match the given graph characteristics, including points $A$ and $B$ where they intersect near $90^\circ$ and $270^\circ$.