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

1 answer

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.