1. Stating the problem: We want to determine the phase shift values (d) for the cosine and sine functions from the given graph. 2. Recall the general forms: - Sine function: $$y = A \sin(B(x - d_{sin})) + C$$ - Cosine function: $$y = A \cos(B(x - d_{cos})) + C$$ where $A$ is amplitude, $B$ relates to period, $C$ is midline, and $d$ is phase shift. 3. From the graph, amplitude $A = 3$ and midline $C = -1$ as calculated. 4. The period $T$ is the length of one full cycle. From the graph, one full cycle spans from $-3\pi/2$ to $3\pi/2$, so: $$T = 3\pi/2 - (-3\pi/2) = 3\pi$$ 5. Calculate $B$ using: $$B = \frac{2\pi}{T} = \frac{2\pi}{3\pi} = \frac{2}{3}$$ 6. To find $d_{sin}$ (phase shift for sine): - The sine function normally starts at zero at $x=0$. - From the graph, sine crosses midline going upward at $x = -3\pi/2$. - So, set inside sine to zero at $x = -3\pi/2$: $$B(x - d_{sin}) = 0 \Rightarrow \frac{2}{3}(-\frac{3\pi}{2} - d_{sin}) = 0$$ 7. Solve for $d_{sin}$: $$\frac{2}{3}(-\frac{3\pi}{2} - d_{sin}) = 0$$ $$-\frac{2}{3} \cdot \frac{3\pi}{2} - \frac{2}{3} d_{sin} = 0$$ $$-\pi - \frac{2}{3} d_{sin} = 0$$ $$- \frac{2}{3} d_{sin} = \pi$$ $$d_{sin} = -\frac{3}{2} \pi$$ 8. To find $d_{cos}$ (phase shift for cosine): - Cosine normally starts at maximum at $x=0$. - From the graph, maximum occurs at $x = -\pi$. - Set inside cosine to zero at $x = -\pi$: $$B(x - d_{cos}) = 0 \Rightarrow \frac{2}{3}(-\pi - d_{cos}) = 0$$ 9. Solve for $d_{cos}$: $$\frac{2}{3}(-\pi - d_{cos}) = 0$$ $$-\frac{2}{3} \pi - \frac{2}{3} d_{cos} = 0$$ $$- \frac{2}{3} d_{cos} = \frac{2}{3} \pi$$ $$d_{cos} = -\pi$$ 10. Final answers: - Phase shift for sine: $$d_{sin} = -\frac{3\pi}{2}$$ - Phase shift for cosine: $$d_{cos} = -\pi$$