6.05 Transformations of sine and cosine functions

f \left( x \right) = 5 \sin x

Graph f \left( x \right)= 5 \sin x with y=\sin x on the same coordinate grid using a table of values, then use the graph to determine its amplitude.

Since the amplitude is half the distance between the maximum and minimum values of a periodic function, we can see that the amplitude of {f \left( x \right) = 5 \sin x} is 5.

We know that the amplitude of the parent function y= \sin x is 1, and the amplitude of {f \left( x \right) = 5 \sin x} is 5. We can see that the value of a in the general sinusoidal form of the function leads to horizontal stretches and compressions, and that the amplitude of the transformed function will be |a|.

The graph of y= \cos x is reflected across the x-axis, then compressed in the vertical direction so that its minimum value is -\dfrac{3}{4}.

Graph the function y=\cos x, then reflect the function across the x-axis, and compress it so that its minimum value is -\dfrac{3}{4}, leading its maximum value to become \dfrac{3}{4}.

Start with the graph of y=\cos x:

Then, reflect y=\cos x across the x-axis:

Finally, compress the reflected graph such that its minimum values become -\dfrac{3}{4} and its maximum values become \dfrac{3}{4}:

The amplitude of the function is \dfrac{3}{4}.

After applying the transformations to the parent function, we can use them to help us build the equation of the function.

We know that the vertical compression changes the value of a, and reflecting across the x-axis also makes that value negative. The equation of the transformed function would be f \left(x \right) = -\dfrac{3}{4} \cos x.

Compare the period, midline, amplitude, and domain and range of g \left( x \right) to the key features of the parent function f \left(x \right) = \sin x.

Graphing the parent function on the same coordinate plane as g \left( x \right) will help us identify changes in the key features.

The midline, amplitude, and domain and range have not changed from the parent function to the transformed function. However, the period has changed from 2 \pi to \pi.

Write the equation of g \left( x \right).

We know that the period of the parent function f \left( x \right) = \sin x is 2 \pi, but the graph of g \left( x \right) is \pi.

If the amplitude, midline, or range of the function were to change, we would also see a change in the values of a and d in the transformed equation.

The graph of g \left(x \right) experiences a change in its period, from 2 \pi to \pi, meaning that the function f \left( x \right) = a \sin \left[b\left( x - c \right)\right] + d must have a change in its b-value, which changes the period to \dfrac{2\pi}{|b|}, so b=2 or b=-2.

We can note that y=\sin \left( -2x \right) would represent a reflection across the y-axis, and we can see by the graph that no reflection occurred. So, b=2 and the equation of the function is g \left(x \right)= \sin2x.

Complete the table of values for both functions:

Substitute each value of x into both f \left(x \right) and g \left( x \right) and evaluate.

Graph f \left(x \right) and g \left(x \right) on the same coordinate plane.

Use the table of values from part (a) to plot the functions on the same coordinate plane and connect the points with a curve.

Describe the transformation from f \left(x \right) to g \left(x \right).

Examine the key features of both graphs and identify any changes.

The amplitude, period and midline remain the same, so the graph has not been stretched or compressed either vertically or horizontally. We can see that the x-intercepts and the minimum and maximum points have all shifted to the right by \dfrac{\pi}{2}, so the transformation is a phase shift to the right \dfrac{\pi}{2} units.

Once we understand how transformations to the graph are related to the general form of the equation, we can recognize the transformation steps by looking at the equation alone.

When we compare g \left( x \right) = \sin \left( x - \dfrac{\pi}{2} \right) to the general equation f \left(x \right) = \sin x, we can see that only the value of c has altered. We know that the phase shift for this equation will be c = \dfrac{\pi}{2}.

Graph the transformed function described.

Apply each transformation to the parent function y= \cos x.

Start with the graph of y=\cos x:

Reflect y=\cos x across the x-axis:

Translate the reflected graph left by \dfrac{\pi}{3} units and down by 3 units:

Determine the equation of the transformed function in the form g \left(x \right) = a\cos \left[b \left( x - c \right)\right] + d, where c is the least positive value in radians.

If the graph of a function is reflected across the x-axis, the resulting function is given by -f\left( x \right), so the value of a=-1.

A horizontal translation or phase shift is given by the value of c and a vertical translation by the value of d. No horizontal compression or change in the period of the function is listed, so we will not change the value of b.

The equation of the function is g \left( x \right) = - \cos \left(x + \dfrac{\pi}{3} \right) - 3.

We can verify that the equation matches the graph of the transformed function using technology:

State the midline, amplitude, period, and domain and range of the function.

Use the graph of the transformed function from part (a) to identify key features:

• Midline: y=-3

• Amplitude: 1

• Period: 2 \pi

• Domain: \left( -\infty, \infty \right)

• Range: [-4,-2]

Explain how the graph and function would be different if we performed the translations before the reflection.

Apply the algebraic translations to the equation of the function first, then the reflection.

The equation of the functions with the translations applied to it would be h \left( x \right) = \cos \left(x + \dfrac{\pi}{3} \right) - 3. Then, if we reflected that function across the x-axis, we would have h \left( x \right) = - \left(\cos \left(x + \dfrac{\pi}{3} \right) - 3\right) = -\cos \left(x + \dfrac{\pi}{3} \right) + 3

The graph of the function would be located in quadrants 1 and 2 as opposed to where the originally transformed function was located, in quadrants 3 and 4.