Question

The function `f(x)=sin(3x)+cos(3x)` is the result of series of transformations with the first one being a horizontal translation of the function `sin(x)`. Which of this describes the first transformation?

Answer 1

We can get the graph of `y=f(x)` from `ysinx` by applying the following transformations:

• a horizontal translation of `pi/12` radians to the left

• a stretch along `Ox` with a scale factor of `1/3` units

• a stretch along `Oy` with a scale factor of `sqrt(2)` units

Explanation:

Consider the function:

`f(x) = sin(3x)+cos(3x)`

Let us suppose we can write this linear combination of sine and cosine as a single phase shifted sine function, that is suppose we have:

`f(x) -= Asin(3x+alpha)`
`\ \ \ \ \ \ \ = A{sin3xcosalpha+cos3xsinalpha}`
`\ \ \ \ \ \ \ = Acosalpha sin3x + Asinalphacos3x`

In which case by comparing coefficients of `sin3x` and `cos3x` we have:

`Acos alpha = 1 \ \ \` and `\ \ \ Asinalpha = 1`

By squaring and adding we have:

`A^2cos^2alpha+A^2sin^2alpha = 2 => A^2=2=> A=sqrt(2)`

By dividing we have:

`tan alpha => alpha=pi/4`

Thus we can write, `f(x)` in the form:

`f(x) -= sin(3x)+cos(3x)`
`\ \ \ \ \ \ \ = sqrt(2)sin(3x+pi/4)`
`\ \ \ \ \ \ \ = sqrt(2)sin(3(x+pi/12))`

So we can get the graph of `y=f(x)` from `ysinx` by applying the following transformations:

• a horizontal translation of `pi/12` radians to the left
• a stretch along `Ox` with a scale factor of `1/3` units
• a stretch along `Oy` with a scale factor of `sqrt(2)` units

Which we can see graphically:

The graph of `y=sinx`:

graph{sinx [-10, 10, -2, 2]}

The graph of `y=sin(x+pi/12)`:

graph{sin(x+pi/12) [-10, 10, -2, 2]}

The graph of `y=sin(3(x+pi/12)) = sin(3x+pi/4)`:

graph{sin(3x+pi/4) [-10, 10, -2, 2]}

The graph of `y=sqrt(2)sin(3(x+pi/12)) = sqrt(2)sin(3x+pi/4)`:

graph{sqrt(2)sin(3x+pi/4) [-10, 10, -2, 2]}

And finally, the graph of the original function for comparison:

graph{sin(3x)+cos(3x) [-10, 10, -2, 2]}