(1) What is the phase shift of `f(x) = 1/2 sin (1/5x - pi) + 3` ? (2) Write the equation of the cosine function with an amplitude of 1, a period of `(3pi)/7` , a phase shift of  `-pi/5` , and a vertical shift of 2 units down.

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The phase shift of a sine function in the form `f(x)=Asin(Bx - C) + D` is `C/B` .

So, to determine the phase shift of `f(x)=1/2sin(1/5x-pi)+3` , plug-in `B=1/5` and `C=pi` into the formula.

Phase Shift `=C/B=pi/(1/5)=5pi`

Hence, the phase shift of the given sine function is `5pi` .

To determine the equation of the cosine function, apply the formula `f(x)=Acos(Bx-C)+D` where A represents the amplitude and D is the vertical shift.

So, plug-in A=1 and D=-2.

`f(x)=1cos(Bx - C)+(-2)`

`f(x) =cos(Bx - C) -2`

To get the value of B, use the formula of period of cosine which is:

Period `=(2pi)/|B|`

So, plug-in Period`=(3pi)/7` .




`B=-14/3, 14/3`

Next, solve for the value of C. Use the formula of phase shift which is:

Phase Shift`=C/B`

So, plug-in Phase Shift `=-1/5`  and `B=-14/3` .




So, when `B=-14/3` , the value of C is `14/15` . If we plug-in these two values to `f(x) =cos(Bx - C) -2` , then the function is:

`f(x)=cos(-14/3x - 14/15) -2`

Also, solve for the other value of C when `B=14/3` .




Hence, when `B=14/3` , the value of C is `-14/15` . So, the other function is:

`f(x)=cos(14/3x + 14/15) -2`

Therefore, there are two cosine functions when its graph has an amplitude of 1,  period of `(3pi)/7` , a phase shift of `-pi/5` , and a vertical shift of 2 units down. These are `f(x)=cos(-14/3x - 14/15) -2` and `f(x)=cos(14/3x + 14/15) -2` .

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