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Write a cosine function that has an amplitude of 5, a midline of 2 and a period of \frac{2\pi}{5} 5 2π ​ .

Sagot :

facundo3141592

Answer:

f(x) = 5*cos(5*x) + 2

Step-by-step explanation:

A general cosine function is written as:

f(x) = A*cos(ω*x) + M

where:

A is the amplitude

M is the midline

ω is the angular frequency.

In this case we know that:

The amplitude is 5, then A = 5

The midline is 2, then M = 2

The period is 2*pi/5

The relation between the angular frequency ω and the period T is:

ω = 2*pi/T

Then if the period is T = 2*pi/5, replacing that in the above equation we find that:

ω = 2*pi/T = 2*pi/(2*pi/5) = 5

ω = 5

Then the function is:

f(x) = 5*cos(5*x) + 2

The cosine function should be [tex]f(x) = 5\times cos(5\times x) + 2[/tex]

  • The calculation is as follows:

A general cosine function is written as:

[tex]f(x) = A\times cos(\omega\times x) + M[/tex]

where

A is the amplitude

M is the midline

ω is the angular frequency.

Here, in the given situation  

The amplitude is 5, then A = 5

The midline is 2, then M = 2

The period is [tex]2\times \pi\div 5[/tex]

The relation between the angular frequency ω and the period T is:

[tex]\omega = 2\times \pi\div T[/tex]

Now Then if the period is [tex]T = 2\div \pi\div 5,[/tex] replacing that in the above equation we find that:

[tex]\omega = 2\times \pi\div T = 2\times \pi\div (2\times \pi\div 5) = 5[/tex]

ω = 5

Then the function is [tex]f(x) = 5\times cos(5\times x) + 2[/tex]

Learn more: brainly.com/question/16911495

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