The function f(x)=sin(3x)+cos(3x)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)sin(x). Which of this describes the first transformation?

enter image source here

1 Answer
Mar 9, 2018

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

  • a horizontal translation of pi/12π12 radians to the left

  • a stretch along OxOx with a scale factor of 1/313 units

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

Explanation:

Consider the function:

f(x) = sin(3x)+cos(3x) 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) f(x)Asin(3x+α)
\ \ \ \ \ \ \ = 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]}