Question

x: -pi/4 pi/4 3pi/4 5pi/4 7pi/2
F(x): 0 1 0 -1 0

These are the values of a sine function at intervals on pi/2. Identify the equation of the function

Answer 1

To identify the equation of the sine function given the values at the specified intervals, let’s analyze the points.

1. **Given points:** We have values of \( F(x) \) at \( x = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{2} \) corresponding to \( F(x) = 0, 1, 0, -1, 0 \).

2. **Identify the pattern:**
- The function reaches its maximum value of \( 1 \) at \( x = \frac{\pi}{4} \).
- The function reaches its minimum value of \( -1 \) at \( x = \frac{5\pi}{4} \).
- The function returns to \( 0 \) at \( x = -\frac{\pi}{4}, \frac{3\pi}{4}, \text{ and } \frac{7\pi}{2} \).

3. **Properties of sine function:**
The sine function has a period of \( 2\pi \) and varies between -1 and 1. The standard form of the sine function is:

\[
F(x) = A \sin(B(x - C)) + D
\]

where:
- \( A \) is the amplitude,
- \( B \) affects the period,
- \( C \) shifts the graph horizontally,
- \( D \) shifts the graph vertically.

4. **Determine amplitude and vertical shift:**
Here, the function oscillates from -1 to 1, so the amplitude \( A = 1 \) and there’s no vertical shift \( D = 0 \). So, we have:

\[
F(x) = \sin(B(x - C))
\]

5. **Find the period:**
We need to find the value of \( B \). The sine function has a standard period of \( 2\pi \). To find this, we can note the transition from \( 1 \) to \( -1 \) in the given values occurs at \( \frac{\pi}{2} \), accordingly:

The sine function achieves a complete cycle every \( 2\pi \). However, we want the period that matches the given points:

Let’s check the period with one full cycle:
- **Between \( -\frac{\pi}{4} \) and \( \frac{7\pi}{2} \):**
- The distance from \( -\frac{\pi}{4} \) to \( \frac{7\pi}{2} \) is \( \frac{7\pi}{2} + \frac{\pi}{4} = \frac{14\pi + \pi}{4} = \frac{15\pi}{4} \).

However, the function cycles fully every \( \pi \) since it returns to a value of 0 at intervals of \( \pi \). So, the period would be \( \pi \):

\[
B = \frac{2\pi}{\pi} = 2
\]

6. **Horizontal shift:**
The function achieves its maximum (1 at \( \frac{\pi}{4} \)), we convert the sine to a cosine to set the phase:

\[
F(x) = \sin\left(2\left(x - \frac{\pi}{4}\right)\right) = \sin(2x - \frac{\pi}{2}) = -\cos(2x)
\]

So, the equation of the function can be represented as:

\[
F(x) = -\cos(2x)
\]

This function will match all given points correctly and returns to zero at \( -\frac{\pi}{4}, 3\frac{\pi}{4}, 7\frac{\pi}{2} \), and reaches \( 1 \) and \( -1 \) at \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \) respectively, confirming the values set in the problem.