Question

find the period of the sine function that has a maximum at $\left(\frac{\pi}{14},1\
ight)$ and a minimum at $\left(\frac{3\pi}{14}, - 1\
ight)$. (1 point)
$\frac{2\pi}{7}$
$\frac{\pi}{14}$
$\frac{3\pi}{14}$
-1

Explanation:

Step1: Determine half - period

The distance between a maximum and a minimum of a sine function is half of its period. Let $x_1=\frac{\pi}{14}$ and $x_2 = \frac{3\pi}{14}$. The half - period $T_{half}=x_2 - x_1$.
$T_{half}=\frac{3\pi}{14}-\frac{\pi}{14}=\frac{3\pi - \pi}{14}=\frac{2\pi}{14}=\frac{\pi}{7}$

Step2: Calculate the period

Since the period $T$ of a sine function is twice the half - period, $T = 2T_{half}$.
$T=2\times\frac{\pi}{7}=\frac{2\pi}{7}$

Answer:

$\frac{2\pi}{7}$