Question

Find the period for a sine function that has a maximum at (pi/14,1) and a minimum at (3pi/14,-1) And write the equation

Answer 1

To find the period of the sine function that has a maximum at \((\frac{\pi}{14}, 1)\) and a minimum at \((\frac{3\pi}{14}, -1)\), we can start by recognizing the properties of the sine wave.

1. **Finding the Period**:
- The maximum occurs at \(\frac{\pi}{14}\).
- The minimum occurs at \(\frac{3\pi}{14}\).

The distance between the maximum and the minimum is:
\[
\frac{3\pi}{14} - \frac{\pi}{14} = \frac{2\pi}{14} = \frac{\pi}{7}.
\]

From a maximum to a minimum is \(\frac{1}{2}\) of a full cycle (since the sine function goes from maximum to minimum). Therefore, the full period \(P\) is:
\[
P = 2 \times \frac{\pi}{7} = \frac{2\pi}{7}.
\]

2. **Writing the Equation**:
The general form of a sine function is:
\[
y = A \sin(B(x - C)) + D,
\]
where:
- \(A\) is the amplitude (1 in this case, as the max is 1 and min is -1),
- \(B\) is related to the period by \(B = \frac{2\pi}{P}\),
- \(C\) is the phase shift,
- \(D\) is the vertical shift.

Given \(P = \frac{2\pi}{7}\):
\[
B = \frac{2\pi}{P} = \frac{2\pi}{\frac{2\pi}{7}} = 7.
\]

Since the sine function has a maximum at \((\frac{\pi}{14}, 1)\), we consider a cosine function starting at maximum for simplicity (as \(\cos(x)\) starts at its maximum). A sine function can be converted by phase shifting:
\[
y = A \cos(B(x - C)) + D.
\]
Here,
- \(A = 1\) (the amplitude),
- \(B = 7\),
- There is no vertical shift (\(D = 0\)),
- The maximum at \(\frac{\pi}{14}\) indicates \(C = \frac{\pi}{14}\).

Thus, the equation becomes:
\[
y = \cos\left(7x - \frac{\pi}{2}\right).
\]

We can also convert back to sine if needed:
\[
y = \sin\left(7x - \frac{3\pi}{2}\right) \quad \text{(equivalent to a cosine function due to phase shift)}.
\]

3. **Final Answer**:
The equation of the sine function is:
\[
y = \sin\left(7x - \frac{3\pi}{2}\right),
\]
or, equivalently,
\[
y = \cos(7x).
\]
The period of the function is \(\frac{2\pi}{7}\).