Answer:
[tex]y = \frac{3}{2}\sin(6x) + \frac{15}{2}[/tex]
Step-by-step explanation:
Given
[tex]T = 60[/tex] --- period
[tex]\{y = ER| 6 \le y \le 9\}[/tex] --- range
Required
The sine function
A sine function is represented as:
[tex]y = A\sin[k(x - d)] + c[/tex]
Where:
[tex]amplitude = |A|[/tex]
and
[tex]T = \frac{360^o}{|k|}[/tex]
[tex]A = \frac{|y_{max} - y_{min}|}{2}[/tex]
and
[tex]c = \frac{y_{min} + y_{max}}{2}[/tex]
Given that the range is:[tex]\{y = ER| 6 \le y \le 9\}[/tex]
This implies that:
[tex]y_{min} = 6[/tex]
[tex]y_{max} = 9[/tex]
So, we have:
[tex]A= \frac{|9-6|}{2}[/tex]
[tex]A= \frac{|3|}{2}[/tex]
Remove absolute bracket
[tex]A= \frac{3}{2}[/tex]
Next, calculate k
[tex]T = \frac{360^o}{|k|}[/tex]
[tex]60 = \frac{360^o}{|k|}[/tex]
Make |k| the subject
[tex]|k| = \frac{360^o}{60}[/tex]
[tex]|k| = 6[/tex]
Remove absolute bracket
[tex]k = 6[/tex]
Next, calculate c
[tex]c = \frac{y_{min} + y_{max}}{2}[/tex]
[tex]c = \frac{6 + 9}{2}[/tex]
[tex]c = \frac{15}{2}[/tex]
So, we have:
[tex]y = A\sin[k(x - d)] + c[/tex]
[tex]y = \frac{3}{2}\sin[6(x -d)] + \frac{15}{2}[/tex]
Set d to 0
[tex]y = \frac{3}{2}\sin(6x) + \frac{15}{2}[/tex]
