Write the equation of the sine function where:

- the amplitude is 5

-the period is `6pi`

-the displacement is `pi/5` units to the left

the main equation to apply the above transformation is:

`y=asin(1/k(x-d))+c`

where

a is vertical stretch (amplitude)

k is horizontal compression

d is translation to the left (-) or right (+)

c is translation up (+) or down (-)

we have amplitude 5 so a=5,

to get k, we know that `(2pi)/k=period` so `(2pi)/k=6pi` solve for k :

`(2pi)/k=6pi` multiply both side by `k` and we have `k*(2pi)/k=6pi*k`

the k on left cancel out and we have `2pi=6pi*k`.

now we have on both side and it cancels out and we have:

`2=6k`

Now divide both side by `6` and we have: `(2)/(6)=(6*k)/(6)`. Now 6 on the right cancel out each other and we have: `2/6=k` simplify, we get:

`k=1/3` and to get `1/(1/3)=3` so we use 3 for the period

to get d, we have `pi/5` to to the left meaning `-pi/5=x` set it =0 and have `0=x+pi/5`

Putting it together we have:

`y=5sin(3(x+pi/5))` and we have simplify the inner parenthesis values and have `y=5sin(3x+(3pi)/5)`

so our final transformed function is: `y=5sin(3x+(3pi)/5)`