Which of the following have a frequency factor of $b=1$?


  • a sine function whose period is $2 \pi$ radians
  • a sine function whose $x$-coefficient is $2 \pi$
  • a cosine function with no phase shift whose $x$-coefficient is 1
  • a sine function whose graph shows 2 cycles from $-4 \pi$ radians to 0
  • a sine function whose graph shows 2 cycles from 0 to $2 \pi$ radians
  • a cosine function whose graph shows 4 cycles from 0 to $4 \pi$ radians
  • a cosine function whose graph shows 1 cycle from $3 \pi$ radians to $5 \pi$ radians

Verified Answer :

  • Identify the relationship between the period \(T\) and the frequency factor \(b\) as \(b = \frac{2\pi}{T}\).
  • Evaluate each option by calculating its period or identifying its \(x\)-coefficient, where \(b=1\) requires \(T=2\pi\) or an \(x\)-coefficient of 1.
  • The following descriptions have \(b=1\):
  • A sine function with a period of \(2\pi\) radians.
  • A cosine function with an \(x\)-coefficient of \(1\).
  • A sine function with \(2\) cycles from \(-4\pi\) to \(0\).
  • A cosine function with \(1\) cycle from \(3\pi\) to \(5\pi\).
  • \(\boxed{\text{Options 1, 3, 4, and 7}}\)

Explanation

  1. Analysis of the Frequency Factor Definition

A trigonometric function of the form \(f(x) = A \sin(b(x - h)) + k\) or \(f(x) = A \cos(b(x - h)) + k\) has a frequency factor \(b\), which determines the period \(T\) of the function through the formula \(T = \frac{2\pi}{|b|}\). Solving for \(b\), we have \(|b| = \frac{2\pi}{T}\). We are looking for functions where the frequency factor \(b = 1\). This is equivalent to finding functions with a period \(T = 2\pi\) or those where the coefficient of \(x\) inside the trigonometric function is \(1\). We will examine each given option to see if it satisfies \(b = 1\).

  1. Evaluation of Each Option

We evaluate each option given the criteria \(b = 1\):

  1. Sine function with period \(T = 2\pi\): Since \(T = \frac{2\pi}{b}\), setting \(T = 2\pi\) gives \(2\pi = \frac{2\pi}{b}\), so \(b = 1\). (Match)
  2. Sine function with \(x\)-coefficient \(2\pi\): The \(x\)-coefficient is the value of \(b\). Here \(b = 2\pi eq 1\). (No Match)
  3. Cosine function with \(x\)-coefficient \(1\): The \(x\)-coefficient is \(b\). Here \(b = 1\). (Match)
  4. Sine function showing \(2\) cycles from \(-4\pi\) to \(0\): The interval length is \(L = 0 - (-4\pi) = 4\pi\). Since there are \(n = 2\) cycles, the period is \(T = \frac{L}{n} = \frac{4\pi}{2} = 2\pi\). As found before, \(T = 2\pi\) implies \(b = 1\). (Match)
  5. Sine function showing \(2\) cycles from \(0\) to \(2\pi\): The interval length is \(L = 2\pi - 0 = 2\pi\). Since there are \(n = 2\) cycles, the period is \(T = \frac{L}{n} = \frac{2\pi}{2} = \pi\). Thus, \(b = \frac{2\pi}{\pi} = 2 eq 1\). (No Match)
  6. Cosine function showing \(4\) cycles from \(0\) to \(4\pi\): The interval length is \(L = 4\pi - 0 = 4\pi\). Since there are \(n = 4\) cycles, the period is \(T = \frac{L}{n} = \frac{4\pi}{4} = \pi\). Thus, \(b = \frac{2\pi}{\pi} = 2 eq 1\). (No Match)
  7. Cosine function showing \(1\) cycle from \(3\pi\) to \(5\pi\): The interval length is \(L = 5\pi - 3\pi = 2\pi\). Since there are \(n = 1\) cycle, the period is \(T = \frac{L}{n} = \frac{2\pi}{1} = 2\pi\). Thus, \(b = 1\). (Match)
  1. Final Conclusion

The options with a frequency factor of \(b=1\) are:

  • A sine function whose period is \(2\pi\) radians.
  • A cosine function with no phase shift whose \(x\)-coefficient is \(1\).
  • A sine function whose graph shows \(2\) cycles from \(-4\pi\) radians to \(0\).
  • A cosine function whose graph shows \(1\) cycle from \(3\pi\) radians to \(5\pi\) radians.

These are the options that satisfy the condition. The final answer is clearly determined.

Examples

The concept of frequency factor is essential in fields like acoustics and signal processing. For instance, in music, the pitch of a sound is determined by its frequency. A sound wave with a frequency factor of \(b=1\) completes a cycle in a specific timeframe (defined by the period), producing a base note. When you adjust the frequency factor of a sound signal on an equalizer, you are effectively stretching or compressing the sound wave, which changes the frequency and thus the pitch, demonstrating how mathematical parameters like \(b\) directly alter physical phenomena.