Does anyone know the equation to this trigonometric function?

A general cosine function (we could also use a sine function) is written as:

y = A*cos(k*x + p) + M

We will find that the function of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

Let's return to the general function:

y = A*cos(k*x + p) + M

A is the amplitude, it defines the distance between the value of a maximum and the value of the minimum, such that A is exactly half of that difference.

Here we can see that the maximum is 0, and the minimum is -4

The differene is: 0 - (-4) = 4

Then:

A = 4/2 = 2

f(x) = 2*cos(k*x + p) + M.

M is the midline, this is, the horizontal line that cuts the graph in two halves. Here we can see that the midline is x = -2, then:

M = -2

f(x) = 2*cos(k*x + p) - 2

p is the phase shift.

In the graph, we can see that f(0) = -3, so we have:

f(0) = 2*cos(0 + p) - 2 = -3

      cos(p) = -1/2

             p = Acos(-1/2) = 2.09

Then we have:

f(x) = 2*cos(k*x + 2.09) - 2

Finally, k is related to the frequency of the function.

We can see that the function does a complete cycle at x = pi

This means that:

f(x) = f(x + pi)

Knowing that the period of a cosine function is 2*pi, then:

k*(x + pi) = k*x + 2*pi

k = 2

Then the equation of the graph is:

f(x) = 2*cos(2*x + 2.09) - 2

If you want to learn more, you can read:

25 : 1

Step-by-step explanation:

The diameter of the football is 5 times the diameter of the cricket ball. This means that the radius of the football is also 5 times the radius of the cricket ball.

Let the radius of the football be R.

Let the radius of the cricket ball be r.

This implies that:

R = 5r

Both balls are spherical.

The surface area of the cricket ball is given as:

The surface area of the football is given as:

Therefore, the ratio of their surface areas is:

A : a

= 100 : 4

25 : 1