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?

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]}