Question: Question 27
Above is a graph of a shifted sinusoidal function. State the function's formulas if the function was originally a regular (positive) sine and cosine function. Give EXACT answers -- no decimals!
As a shifted sine function: y =
As a shifted cosine function: y =
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graph: a blue sinusoidal curve centered on y = 0 with amplitude 2 and period 7\pi/2, crossing the x-axis at x = -7/4, 7/4, 21/4, and 35/4; position_hint = center
1. **State the problem:**
We have a sinusoidal function with amplitude $2$, period $\frac{7\pi}{2}$, and it crosses the x-axis at $x = -\frac{7}{4}, \frac{7}{4}, \frac{21}{4}, \frac{35}{4}$. We want to find the exact formulas for this function as a shifted sine and as a shifted cosine.
2. **Recall the general forms:**
- For sine: $$y = A \sin\left(B(x - C)\right) + D$$
- For cosine: $$y = A \cos\left(B(x - C)\right) + D$$
where:
- $A$ is amplitude
- $B = \frac{2\pi}{\text{period}}$
- $C$ is horizontal shift
- $D$ is vertical shift
3. **Identify given values:**
- Amplitude $A = 2$
- Period $= \frac{7\pi}{2}$
- Vertical shift $D = 0$ (centered on $y=0$)
Calculate $B$:
$$B = \frac{2\pi}{\frac{7\pi}{2}} = \frac{2\pi}{7\pi/2} = \frac{2\pi \times 2}{7\pi} = \frac{4}{7}$$
4. **Find horizontal shift $C$ for sine:**
Sine normally crosses zero at $x=0$. Here, the first zero crossing is at $x = -\frac{7}{4}$.
So,
$$B(x - C) = 0 \implies x - C = 0 \implies x = C$$
Since zero crossing is at $x = -\frac{7}{4}$,
$$C = -\frac{7}{4}$$
5. **Write the sine function:**
$$y = 2 \sin\left(\frac{4}{7}\left(x + \frac{7}{4}\right)\right)$$
6. **Find horizontal shift $C$ for cosine:**
Cosine normally has a maximum at $x=0$. The graph crosses zero at $x = -\frac{7}{4}$, so the maximum is a quarter period before that zero crossing.
Quarter period:
$$\frac{1}{4} \times \frac{7\pi}{2} = \frac{7\pi}{8}$$
Since $B = \frac{4}{7}$, the phase shift $C$ satisfies:
$$B C = \text{phase shift in radians}$$
We want the maximum at
$$x = -\frac{7}{4} - \frac{7\pi}{8} \times \frac{1}{B}$$
But easier is to note cosine leads sine by $\frac{\pi}{2}$ in phase:
$$\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$$
So,
$$y = 2 \cos\left(\frac{4}{7}x + \phi\right)$$
We find $\phi$ by shifting sine phase by $-\frac{\pi}{2}$:
From sine:
$$\frac{4}{7}\left(x + \frac{7}{4}\right) = \frac{4}{7}x + 1$$
(since $\frac{4}{7} \times \frac{7}{4} = 1$)
So sine is:
$$y = 2 \sin\left(\frac{4}{7}x + 1\right)$$
Cosine is:
$$y = 2 \cos\left(\frac{4}{7}x + 1 - \frac{\pi}{2}\right)$$
7. **Final answers:**
- Shifted sine function:
$$y = 2 \sin\left(\frac{4}{7}x + 1\right)$$
- Shifted cosine function:
$$y = 2 \cos\left(\frac{4}{7}x + 1 - \frac{\pi}{2}\right)$$