Question

Determine the equation of a sine function that would have a range of 6 <=y<=9 and a period of 60

Answer 1

`y = (3)/(2)\sin(6x) + (15)/(2)`

Explanation:

Given

`T = 60` --- period

`\y = ER` --- range

Required

The sine function

A sine function is represented as:

`y = A\sin[k(x - d)] + c`

Where:

`amplitude = |A|`

and

`T = (360^o)/(|k|)`

`A = (|y_(max) - y_(min)|)/(2)`

and

`c = (y_(min) + y_(max))/(2)`

Given that the range is:
`\ 6 \le y \le 9\`

This implies that:

`y_(min) = 6`

`y_(max) = 9`

So, we have:

`A= (|9-6|)/(2)`

`A= (|3|)/(2)`

Remove absolute bracket

`A= (3)/(2)`

Next, calculate k

`T = (360^o)/(|k|)`

`60 = (360^o)/(|k|)`

Make |k| the subject

`|k| = (360^o)/(60)`

`|k| = 6`

Remove absolute bracket

`k = 6`

Next, calculate c

`c = (y_(min) + y_(max))/(2)`

`c = (6 + 9)/(2)`

`c = (15)/(2)`

So, we have:

`y = A\sin[k(x - d)] + c`

`y = (3)/(2)\sin[6(x -d)] + (15)/(2)`

Set d to 0

`y = (3)/(2)\sin(6x) + (15)/(2)`