1. **Problem:** Determine the period, amplitude, phase shift, and horizontal shift of the function $f(x) = 4 \sin(4x - 3\pi)$. 2. **Formula and rules:** For a function of the form $f(x) = A \sin(Bx - C)$ or $f(x) = A \cos(Bx - C)$, - Amplitude = $|A|$ - Period = $\frac{2\pi}{|B|}$ - Phase shift = $\frac{C}{B}$ - Horizontal shift is the same as phase shift but direction depends on sign inside the function. 3. **Apply to $f(x) = 4 \sin(4x - 3\pi)$:** - Amplitude = $|4| = 4$ - Period = $\frac{2\pi}{4} = \frac{\pi}{2}$ - Phase shift = $\frac{3\pi}{4}$ - Since inside is $(4x - 3\pi)$, the shift is to the right by $\frac{3\pi}{4}$. --- 1. **Problem:** Determine the period, amplitude, phase shift, and horizontal shift of the function $f(x) = -2 \cos\left(\frac{x}{2} - \frac{\pi}{4}\right) + 1$. 2. **Rewrite inside argument:** $\frac{x}{2} - \frac{\pi}{4} = \frac{1}{2}x - \frac{\pi}{4}$, so $B = \frac{1}{2}$ and $C = \frac{\pi}{4}$. 3. **Apply formulas:** - Amplitude = $|-2| = 2$ - Period = $\frac{2\pi}{|1/2|} = 2\pi \times 2 = 4\pi$ - Phase shift = $\frac{\pi/4}{1/2} = \frac{\pi}{4} \times 2 = \frac{\pi}{2}$ - Shift is to the right by $\frac{\pi}{2}$. - Vertical shift = $+1$ (upwards). --- 1. **Problem:** Determine the period, amplitude, phase shift, and horizontal shift of the function $f(x) = 3 \sin\left(\frac{\pi}{2} - \frac{x}{3}\right) - 1$. 2. **Rewrite inside argument:** $\frac{\pi}{2} - \frac{x}{3} = -\frac{1}{3}x + \frac{\pi}{2}$, so $B = -\frac{1}{3}$ and $C = \frac{\pi}{2}$. 3. **Apply formulas:** - Amplitude = $|3| = 3$ - Period = $\frac{2\pi}{| -1/3 |} = 2\pi \times 3 = 6\pi$ - Phase shift = $\frac{\pi/2}{-1/3} = \pi/2 \times (-3) = -\frac{3\pi}{2}$ - Shift is to the left by $\frac{3\pi}{2}$. - Vertical shift = $-1$ (downwards). --- **Summary for Part 1:** 1. $f(x) = 4 \sin(4x - 3\pi)$: Amplitude = 4, Period = $\frac{\pi}{2}$, Phase shift = $\frac{3\pi}{4}$ right, No vertical shift. 2. $f(x) = -2 \cos\left(\frac{x}{2} - \frac{\pi}{4}\right) + 1$: Amplitude = 2, Period = $4\pi$, Phase shift = $\frac{\pi}{2}$ right, Vertical shift = 1 up. 3. $f(x) = 3 \sin\left(\frac{\pi}{2} - \frac{x}{3}\right) - 1$: Amplitude = 3, Period = $6\pi$, Phase shift = $\frac{3\pi}{2}$ left, Vertical shift = 1 down. --- **Part 2: Complete the table** | rev | deg | rad | |-----|-----|-----| | ? | 1260° | ? | | -9/8 | ? | ? | | ? | ? | $\frac{5\pi}{12}$ | 1. Convert 1260° to revolutions and radians: - Revolutions = $\frac{1260}{360} = 3.5$ rev - Radians = $1260 \times \frac{\pi}{180} = 7\pi$ rad 2. Convert $-\frac{9}{8}$ rev to degrees and radians: - Degrees = $-\frac{9}{8} \times 360 = -405°$ - Radians = $-\frac{9}{8} \times 2\pi = -\frac{9\pi}{4}$ 3. Convert $\frac{5\pi}{12}$ rad to degrees and revolutions: - Degrees = $\frac{5\pi}{12} \times \frac{180}{\pi} = 75°$ - Revolutions = $\frac{5\pi}{12} \times \frac{1}{2\pi} = \frac{5}{24}$ rev --- **Part 3: Compute other five trig functions given:** 1. Given $\cos \theta = \frac{3}{5}$ and $\tan \theta < 0$. - Since $\cos \theta > 0$ and $\tan \theta < 0$, $\theta$ is in quadrant IV. - Use identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\sin \theta$: $$\sin \theta = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$ - $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-4/5}{3/5} = -\frac{4}{3}$ - $\sec \theta = \frac{1}{\cos \theta} = \frac{5}{3}$ - $\csc \theta = \frac{1}{\sin \theta} = -\frac{5}{4}$ - $\cot \theta = \frac{1}{\tan \theta} = -\frac{3}{4}$ 2. Given $\sin \theta = \frac{\sqrt{3}}{3}$ and $\theta \in \left(\frac{\pi}{2}, \pi\right)$. - $\theta$ is in quadrant II where $\sin > 0$, $\cos < 0$, $\tan < 0$. - Use $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$: $$\cos \theta = -\sqrt{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = -\sqrt{1 - \frac{3}{9}} = -\sqrt{\frac{6}{9}} = -\frac{\sqrt{6}}{3}$$ - $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{3}}{3}}{-\frac{\sqrt{6}}{3}} = -\frac{\sqrt{3}}{\sqrt{6}} = -\sqrt{\frac{1}{2}} = -\frac{\sqrt{2}}{2}$ - $\sec \theta = \frac{1}{\cos \theta} = -\frac{3}{\sqrt{6}} = -\frac{3\sqrt{6}}{6} = -\frac{\sqrt{6}}{2}$ - $\csc \theta = \frac{1}{\sin \theta} = \frac{3}{\sqrt{3}} = \sqrt{3}$ - $\cot \theta = \frac{1}{\tan \theta} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$ --- **Final answers:** - Part 1: See detailed amplitude, period, phase shift, and vertical shift for each function. - Part 2: Table completed with conversions. - Part 3: All five trig functions computed for each given condition.