What is the period of the sinusoid given by [tex]y=-2 \sin \left(\frac{2 \pi}{7} x\right)[/tex]?

Answer here: ___________________



Answer :

To determine the period of the sinusoid given by the equation [tex]\( y = -2 \sin \left(\frac{2 \pi}{7} x \right) \)[/tex], we start by recognizing the general form of a sinusoidal function:
[tex]\[ y = A \sin(Bx + C) + D \][/tex]

In this specific case, the equation is:
[tex]\[ y = -2 \sin \left(\frac{2 \pi}{7} x \right) \][/tex]

Here, we can identify the following components:
- The amplitude [tex]\( A \)[/tex] is [tex]\(-2\)[/tex], but the sign and amplitude do not affect the period.
- The coefficient [tex]\( B \)[/tex], multiplying [tex]\( x \)[/tex], is [tex]\(\frac{2 \pi}{7}\)[/tex].

The period [tex]\( T \)[/tex] of a sinusoidal function of the form [tex]\( y = A \sin(Bx + C) + D \)[/tex] can be determined by the formula:
[tex]\[ T = \frac{2 \pi}{|B|} \][/tex]

Substituting the value of [tex]\( B = \frac{2 \pi}{7} \)[/tex]:

[tex]\[ T = \frac{2 \pi}{\left| \frac{2 \pi}{7} \right|} \][/tex]

Since [tex]\(\left| \frac{2 \pi}{7} \right| = \frac{2 \pi}{7}\)[/tex], we have:

[tex]\[ T = \frac{2 \pi}{\frac{2 \pi}{7}} \][/tex]

To simplify this, we multiply by the reciprocal of the denominator:

[tex]\[ T = 2 \pi \cdot \frac{7}{2 \pi} \][/tex]

The [tex]\( 2 \pi \)[/tex] in the numerator and denominator cancel out, leaving:

[tex]\[ T = 7 \][/tex]

Therefore, the period of the sinusoid [tex]\( y = -2 \sin \left( \frac{2 \pi}{7} x \right) \)[/tex] is:
[tex]\[ \boxed{7} \][/tex]