which of the following have a frequency factor of b = 1?\n□ a sine function whose period is 2π radians\n□ a…

which of the following have a frequency factor of b = 1?\n□ a sine function whose period is 2π radians\n□ a sine function whose x - coefficient is 2π\n□ a cosine function with no phase shift whose x - coefficient is 1\n□ a sine function whose graph shows 2 cycles from - 4π radians to 0\n□ a sine function whose graph shows 2 cycles from 0 to 2π radians\n□ a cosine function whose graph shows 4 cycles from 0 to 4π radians\n□ a cosine function whose graph shows 1 cycle from 3π radians to 5π radians

which of the following have a frequency factor of b = 1?\n□ a sine function whose period is 2π radians\n□ a sine function whose x - coefficient is 2π\n□ a cosine function with no phase shift whose x - coefficient is 1\n□ a sine function whose graph shows 2 cycles from - 4π radians to 0\n□ a sine function whose graph shows 2 cycles from 0 to 2π radians\n□ a cosine function whose graph shows 4 cycles from 0 to 4π radians\n□ a cosine function whose graph shows 1 cycle from 3π radians to 5π radians

Answer

Explanation:

Step1: Recall period - frequency formula

The period $T$ of a sinusoidal function $y = A\sin(bx - c)+d$ or $y=A\cos(bx - c)+d$ is given by $T=\frac{2\pi}{|b|}$. When $b = 1$, $T = 2\pi$.

Step2: Analyze each option

Option 1:

A sine - function whose period is $2\pi$ radians. Using $T=\frac{2\pi}{|b|}$, if $T = 2\pi$, then $2\pi=\frac{2\pi}{|b|}$, solving for $b$ gives $|b| = 1$, so $b=\pm1$. This option is correct.

Option 2:

A sine - function whose $x$ - coefficient is $2\pi$. Here $b = 2\pi\neq1$, so this option is incorrect.

Option 3:

A cosine function with no phase - shift whose $x$ - coefficient is $1$. Since the $x$ - coefficient is $b$ and $b = 1$, this option is correct.

Option 4:

A sine function whose graph shows 2 cycles from $-4\pi$ to $0$. The period $T=\frac{0-(-4\pi)}{2}=2\pi$. Using $T = \frac{2\pi}{|b|}$, we get $2\pi=\frac{2\pi}{|b|}$, so $|b| = 1$, $b=\pm1$. This option is correct.

Option 5:

A sine function whose graph shows 2 cycles from $0$ to $2\pi$. The period $T=\frac{2\pi - 0}{2}=\pi$. Using $T=\frac{2\pi}{|b|}$, we have $\pi=\frac{2\pi}{|b|}$, so $|b| = 2$, $b=\pm2$. This option is incorrect.

Option 6:

A cosine function whose graph shows 4 cycles from $0$ to $4\pi$. The period $T=\frac{4\pi - 0}{4}=\pi$. Using $T=\frac{2\pi}{|b|}$, we get $\pi=\frac{2\pi}{|b|}$, so $|b| = 2$, $b=\pm2$. This option is incorrect.

Option 7:

A cosine function whose graph shows 1 cycle from $3\pi$ to $5\pi$. The period $T=5\pi - 3\pi = 2\pi$. Using $T=\frac{2\pi}{|b|}$, we have $2\pi=\frac{2\pi}{|b|}$, so $|b| = 1$, $b=\pm1$. This option is correct.

Answer:

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.