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Write a sine function that has an amplitude of 5, a midline of 2 and a period of 8pi/5.

User Littleadv
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1 Answer

5 votes

Answer:

f(x) = 5*sin( (5/4)*x) + 2

Explanation:

A general sine function can be written as:

f(x) = A*sin(w*x) + M

Where:

A is the amplitude

w is the frequency of the function

M is the midline

Here we know that we want to have:

An amplitude of 5. Then, A = 5

A midline of 2, then M = 2

And a period of (8/5)*pi

For now, we can write our function as:

f(x) = 5*sin(w*x) + 2

Now, we know that for a function of period T, is always true that:

f(x) = f(x + T)

here, the period is T = (8/5)*pi

knowing that sin(0) = sin(2*pi)

Then we must have:

sin(0) = sin(w*0) = sin(w*( 0 + (8/5)*pi)) = sin(2*pi)

Then the arguments of the third and fourt parts must be the same, thus:

w*(8/5)*pi = 2*pi

We can solve this for w now:

w*(8/5) = 2

w = 2*(5/8) = (5/4)

Then the function is:

f(x) = 5*sin( (5/4)*x) + 2

User Fakeaccount
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