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To determine which functions have a frequency factor [tex]\(b = 1\)[/tex], we need to understand how the frequency factor relates to different trigonometric functions. In general, for a trigonometric function of the form [tex]\( f(x) = \sin(bx) \)[/tex] or [tex]\( f(x) = \cos(bx) \)[/tex], the frequency factor [tex]\(b\)[/tex] determines the number of cycles the function completes in a given interval, specifically, a period of [tex]\( \frac{2\pi}{b} \)[/tex].
Let's analyze each option:
1. A sine function whose period is [tex]\(2\pi\)[/tex] radians:
- The period [tex]\(T\)[/tex] of a sine function can be given by [tex]\(T = \frac{2\pi}{b}\)[/tex].
- If the period is [tex]\(2\pi\)[/tex], then [tex]\(\frac{2\pi}{b} = 2\pi\)[/tex].
- Solving for [tex]\(b\)[/tex], we get [tex]\(b = 1\)[/tex].
2. A sine function whose [tex]\(x\)[/tex]-coefficient is [tex]\(2\pi\)[/tex]:
- For a sine function of the form [tex]\(\sin(bx)\)[/tex], the [tex]\(x\)[/tex]-coefficient is [tex]\(b\)[/tex].
- Here, the [tex]\(x\)[/tex]-coefficient is [tex]\(2\pi\)[/tex], so [tex]\(b = 2\pi\)[/tex].
3. A cosine function with no phase shift whose [tex]\(x\)[/tex]-coefficient is 1:
- For a cosine function of the form [tex]\(\cos(bx)\)[/tex], the [tex]\(x\)[/tex]-coefficient is [tex]\(b\)[/tex].
- Here, the [tex]\(x\)[/tex]-coefficient is [tex]\(1\)[/tex], so [tex]\(b = 1\)[/tex].
4. A sine function whose graph shows 2 cycles from [tex]\(-4\pi\)[/tex] radians to 0:
- The interval length is [tex]\(4\pi\)[/tex] (from [tex]\(-4\pi\)[/tex] to 0).
- If there are 2 cycles in [tex]\(4\pi\)[/tex] radians, then each cycle is [tex]\(\frac{4\pi}{2} = 2\pi\)[/tex].
- A period of [tex]\(2\pi\)[/tex] corresponds to [tex]\(b = 1\)[/tex].
5. A sine function whose graph shows 2 cycles from 0 to [tex]\(2\pi\)[/tex] radians:
- The interval length is [tex]\(2\pi\)[/tex] (from 0 to [tex]\(2\pi\)[/tex]).
- If there are 2 cycles in [tex]\(2\pi\)[/tex] radians, then each cycle is [tex]\(\frac{2\pi}{2} = \pi\)[/tex].
- A period of [tex]\(\pi\)[/tex] corresponds to [tex]\(b = \frac{2\pi}{\pi} = 2\)[/tex].
6. A cosine function whose graph shows 4 cycles from 0 to [tex]\(4\pi\)[/tex] radians:
- The interval length is [tex]\(4\pi\)[/tex] (from 0 to [tex]\(4\pi\)[/tex]).
- If there are 4 cycles in [tex]\(4\pi\)[/tex] radians, then each cycle is [tex]\(\frac{4\pi}{4} = \pi\)[/tex].
- A period of [tex]\(\pi\)[/tex] corresponds to [tex]\(b = \frac{2\pi}{\pi} = 2\)[/tex].
7. A cosine function whose graph shows 1 cycle from [tex]\(3\pi\)[/tex] radians to [tex]\(5\pi\)[/tex] radians:
- The interval length is [tex]\(2\pi\)[/tex] (from [tex]\(3\pi\)[/tex] to [tex]\(5\pi\)[/tex]).
- If there is 1 cycle in [tex]\(2\pi\)[/tex] radians, then each cycle is [tex]\(2\pi\)[/tex].
- A period of [tex]\(2\pi\)[/tex] corresponds to [tex]\(b = 1\)[/tex].
Based on this analysis, the functions that have a frequency factor [tex]\(b = 1\)[/tex] are:
- A sine function whose period is [tex]\(2\pi\)[/tex] radians
- A cosine function with no phase shift whose [tex]\(x\)[/tex]-coefficient is 1
- A sine function whose graph shows 2 cycles from [tex]\(-4\pi\)[/tex] radians to 0
- A cosine function whose graph shows 1 cycle from [tex]\(3\pi\)[/tex] radians to [tex]\(5\pi\)[/tex] radians
Let's analyze each option:
1. A sine function whose period is [tex]\(2\pi\)[/tex] radians:
- The period [tex]\(T\)[/tex] of a sine function can be given by [tex]\(T = \frac{2\pi}{b}\)[/tex].
- If the period is [tex]\(2\pi\)[/tex], then [tex]\(\frac{2\pi}{b} = 2\pi\)[/tex].
- Solving for [tex]\(b\)[/tex], we get [tex]\(b = 1\)[/tex].
2. A sine function whose [tex]\(x\)[/tex]-coefficient is [tex]\(2\pi\)[/tex]:
- For a sine function of the form [tex]\(\sin(bx)\)[/tex], the [tex]\(x\)[/tex]-coefficient is [tex]\(b\)[/tex].
- Here, the [tex]\(x\)[/tex]-coefficient is [tex]\(2\pi\)[/tex], so [tex]\(b = 2\pi\)[/tex].
3. A cosine function with no phase shift whose [tex]\(x\)[/tex]-coefficient is 1:
- For a cosine function of the form [tex]\(\cos(bx)\)[/tex], the [tex]\(x\)[/tex]-coefficient is [tex]\(b\)[/tex].
- Here, the [tex]\(x\)[/tex]-coefficient is [tex]\(1\)[/tex], so [tex]\(b = 1\)[/tex].
4. A sine function whose graph shows 2 cycles from [tex]\(-4\pi\)[/tex] radians to 0:
- The interval length is [tex]\(4\pi\)[/tex] (from [tex]\(-4\pi\)[/tex] to 0).
- If there are 2 cycles in [tex]\(4\pi\)[/tex] radians, then each cycle is [tex]\(\frac{4\pi}{2} = 2\pi\)[/tex].
- A period of [tex]\(2\pi\)[/tex] corresponds to [tex]\(b = 1\)[/tex].
5. A sine function whose graph shows 2 cycles from 0 to [tex]\(2\pi\)[/tex] radians:
- The interval length is [tex]\(2\pi\)[/tex] (from 0 to [tex]\(2\pi\)[/tex]).
- If there are 2 cycles in [tex]\(2\pi\)[/tex] radians, then each cycle is [tex]\(\frac{2\pi}{2} = \pi\)[/tex].
- A period of [tex]\(\pi\)[/tex] corresponds to [tex]\(b = \frac{2\pi}{\pi} = 2\)[/tex].
6. A cosine function whose graph shows 4 cycles from 0 to [tex]\(4\pi\)[/tex] radians:
- The interval length is [tex]\(4\pi\)[/tex] (from 0 to [tex]\(4\pi\)[/tex]).
- If there are 4 cycles in [tex]\(4\pi\)[/tex] radians, then each cycle is [tex]\(\frac{4\pi}{4} = \pi\)[/tex].
- A period of [tex]\(\pi\)[/tex] corresponds to [tex]\(b = \frac{2\pi}{\pi} = 2\)[/tex].
7. A cosine function whose graph shows 1 cycle from [tex]\(3\pi\)[/tex] radians to [tex]\(5\pi\)[/tex] radians:
- The interval length is [tex]\(2\pi\)[/tex] (from [tex]\(3\pi\)[/tex] to [tex]\(5\pi\)[/tex]).
- If there is 1 cycle in [tex]\(2\pi\)[/tex] radians, then each cycle is [tex]\(2\pi\)[/tex].
- A period of [tex]\(2\pi\)[/tex] corresponds to [tex]\(b = 1\)[/tex].
Based on this analysis, the functions that have a frequency factor [tex]\(b = 1\)[/tex] are:
- A sine function whose period is [tex]\(2\pi\)[/tex] radians
- A cosine function with no phase shift whose [tex]\(x\)[/tex]-coefficient is 1
- A sine function whose graph shows 2 cycles from [tex]\(-4\pi\)[/tex] radians to 0
- A cosine function whose graph shows 1 cycle from [tex]\(3\pi\)[/tex] radians to [tex]\(5\pi\)[/tex] radians
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