Subjects trigonometry

Sinusoidal Equation 1Bda61

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Sinusoidal Equation 1Bda61


1. **State the problem:** We need to find the equation of a sinusoidal function based on the given graph description. 2. **Identify key features from the graph:** - The sinusoid starts near $y=1$ at $x=-\frac{\pi}{24}$. - It reaches a minimum near $y=-4$ between $x=\frac{\pi}{24}$ and $x=\frac{\pi}{12}$. - It then rises to above $y=5$ slightly before $x=\frac{\pi}{8}$. 3. **Determine amplitude ($A$):** - The maximum value is about $5$ and the minimum is about $-4$. - Amplitude $A = \frac{\text{max} - \text{min}}{2} = \frac{5 - (-4)}{2} = \frac{9}{2} = 4.5$. 4. **Determine vertical shift ($D$):** - Vertical shift $D = \frac{\text{max} + \text{min}}{2} = \frac{5 + (-4)}{2} = \frac{1}{2} = 0.5$. 5. **Determine period ($T$):** - The sinusoid goes from near a midline point at $x=-\frac{\pi}{24}$, to a minimum between $\frac{\pi}{24}$ and $\frac{\pi}{12}$, and back to a maximum near $\frac{\pi}{8}$. - The distance between minimum and maximum is roughly $\frac{\pi}{8} - \frac{\pi}{12} = \frac{3\pi}{24} - \frac{2\pi}{24} = \frac{\pi}{24}$. - Since half a period corresponds to the distance between minimum and maximum, half period $= \frac{\pi}{24}$. - Therefore, full period $T = 2 \times \frac{\pi}{24} = \frac{\pi}{12}$. 6. **Calculate angular frequency ($\omega$):** - $\omega = \frac{2\pi}{T} = \frac{2\pi}{\frac{\pi}{12}} = 24$. 7. **Determine phase shift ($\phi$):** - The sinusoid starts near the midline at $x=-\frac{\pi}{24}$ and is decreasing, which matches the behavior of a cosine shifted to the right. - Using the cosine form: $y = A \cos(\omega x + \phi) + D$. - At $x = -\frac{\pi}{24}$, $y \approx 1$. - Substitute known values: $$1 = 4.5 \cos\left(24 \times \left(-\frac{\pi}{24}\right) + \phi\right) + 0.5$$ $$1 = 4.5 \cos(-\pi + \phi) + 0.5$$ $$1 - 0.5 = 4.5 \cos(-\pi + \phi)$$ $$0.5 = 4.5 \cos(-\pi + \phi)$$ $$\cos(-\pi + \phi) = \frac{0.5}{4.5} = \frac{1}{9}$$ - Since $\cos(-\pi + \phi) = -\cos(\phi)$, we have: $$-\cos(\phi) = \frac{1}{9} \implies \cos(\phi) = -\frac{1}{9}$$ - So $\phi = \arccos\left(-\frac{1}{9}\right)$. 8. **Final equation:** $$ y = 4.5 \cos\left(24x + \arccos\left(-\frac{1}{9}\right)\right) + 0.5 $$ This equation models the sinusoidal function described by the graph.