Finding Phi in Sinusoidal

[JRoullier]

I have every component except phi and I have no clue how to calculate phi.
The equation I currently have is y=4.7sin(4pi/25(x-phi))+4.1
Any help is appreciated thank you.

 

[tkhunny]

What happens when phi is zero?

sin(something) = 0, what is the something?

Keep in mind that there are many possible solutions. Typically, one picks the one closest to zero? You may have instructions to the contrary.

 

[HallsofIvy]

I have every component except phi and I have no clue how to calculate phi.
The equation I currently have is y=4.7sin(4pi/25(x-phi))+4.1
Any help is appreciated thank you.

What you have written is ambiguous. You probably mean "y= 4.7 sin((4pi/25)(x- phi))+ 4.1" but what you wrote could also mean "y= 4.7 sin(4pi/(25(x- phi))+ 4.1"

"Undo" what is done step by step. Assuming you meant the first above, from y= 4.7 sin(4pi/25)(x- phi)+ 4.1 subtract 4.1 from both sides to get y- 4.1= 4.7 sin((4pi/25)(x- phi)). Now, divide both sides by 4.7 to get (y- 4.1)/4.7= sin((4pi/25)(x- phi)). Undo the "sine" by using the inverse function, "arcsin" (which is multi-valued, giving the "many possible values" tkhunny mentioned) : arcsin((y- 4.1)/4.7)=
(4pi/25)(x- phi). Divide both sides by 4pi/25 which is the same as multiplying by 25/(4pi): (25/(4pi))arcsin((y- 4.1)/4.7)= x- phi.

Finally, add phi to both sides: x= phi+ (25/(4pi))arcsin((y- 4.1)/4.7).

 

[Dr.Peterson]

I have every component except phi and I have no clue how to calculate phi.
The equation I currently have is y=4.7sin(4pi/25(x-phi))+4.1
Any help is appreciated thank you.

You haven't stated the entire problem. Are you given a graph of a sinusoid, and trying to write an equation for it? If so, then phi is just the horizontal shift of the graph: the value of x at which the first cycle of the sine starts.

Please show us the original problem and your work, if you need further help, so we can be sure what you are trying to do, and that what you have done so far is correct.