1
Kyran O'Brien
Original investigation was submitted on April 5th, 2022
This report was submitted on May 29th, 2022
Report On Transformations Of Sine Functions And Their Effect On These Functions' Amplitude
And Period
Introduction
We were examining the function 𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷 with regards to the effect of
the constants on the graph of 𝑓 relative to the graph of 𝑦 = 𝑠𝑖𝑛 𝑥 and the amplitude and
period of 𝑓. For a mathematical function, amplitude is
𝑚𝑎𝑥−𝑚𝑖𝑛
2
, while period is the magnitude
of 𝑥-values over which the 𝑦-values of a function repeat themselves. In the case of 𝑦 = 𝑠𝑖𝑛 𝑥,
its amplitude is
1−(−1)
2
= 1 and its period is 2π radians, because that is how many unique
angles there are (one full revolution of a circle) and the sine function's input is always an angle.
Figure 1: This amplitude and period is shown in the hand-drawn graph of 𝑦 = 𝑠𝑖𝑛 𝑥 I made
above to fully anticipate the big picture of where all the ordered pairs on its graph will transfer
to with the application of 𝐴, 𝐵, 𝐶, and 𝐷. As such, points that make the scale unclear and
excessively cluttered - having 𝑥-values with multiples of
𝑦-values of
2
2
π
3
and
π
6
and similarly off-center
- are not included. In this graph, both the 𝑥- and 𝑦- axes have the same,
consistent scale. Likewise, the 𝑦-values for multiples of
explicitly labeled.
π
3
and
π
6
are roughly accurate, just not
2
The role of the constant 𝐴 in 𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷
Figure 2: I used Desmos to graph 𝑓 for different values of 𝐴, ignoring the other constants.
I notice 𝐴 multiplies every 𝑦-value of 𝑦 = 𝑠𝑖𝑛 𝑥 by a factor of 𝐴, stretching the graph vertically
and reflecting it in the 𝑥-axis for 𝐴 < 0 as the 𝑦-values switch signs in this case due to the
multiplication by − 1.
𝐴 affects the amplitude of 𝑦 = 𝑠𝑖𝑛 𝑥 by factor |𝐴| (as amplitude is positive) as well since
𝐴(1)−𝐴(−1)
2
=
𝐴(1−(−1))
2
= 𝐴(
1−(−1)
2
)
however because the amplitude of 𝑦 = 𝑠𝑖𝑛 𝑥 is 1, |𝐴| is the amplitude of 𝑓:
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 = |𝐴|
𝐴 does not affect the period of 𝑓 because it does not affect 𝑥-values (and obviously does not
affect 𝑦-values in different ways over magnitudes greater than the period).
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The role of the constant 𝐵 in 𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷
Figure 3: I graphed 𝑓 for different values of 𝐵, ignoring the other constants.
I notice 𝐵 multiplies every 𝑥-value of 𝑦 = 𝑠𝑖𝑛 𝑥 by a factor of
1
𝐵
because 𝑥-values of
1
𝐵
times
the 𝑥-values of 𝑦 = 𝑠𝑖𝑛 𝑥 are needed in 𝑦 = 𝑠𝑖𝑛(𝐵𝑥) to achieve the same 𝑦-values as
1
𝑦 = 𝑠𝑖𝑛 𝑥: 𝑦 = 𝑠𝑖𝑛(𝐵( 𝐵 𝑥)) = 𝑠𝑖𝑛 𝑥. This stretches the graph horizontally and reflects it in
the 𝑦-axis for 𝐵 < 0 because now 𝑥-values of opposite signs are required to produce the same
𝑦-values as 𝑦 = 𝑠𝑖𝑛 𝑥.
𝐵 does not affect the amplitude because it does not affect 𝑦-values.
𝐵 does affect the period. Because 𝑥-values of
1
𝐵
times the 𝑥-values of 𝑦 = 𝑠𝑖𝑛 𝑥 are needed in
𝑦 = 𝑠𝑖𝑛(𝐵𝑥) to achieve the same 𝑦-values, 𝑦-values on 𝑦 = 𝑠𝑖𝑛(𝐵𝑥) repeat themselves by
times the magnitude of 𝑥-values over which they repeat themselves on 𝑦 = 𝑠𝑖𝑛 𝑥; in other
words, the period of 𝑦 = 𝑠𝑖𝑛(𝐵𝑥) is
1
𝐵
times the period of 𝑦 = 𝑠𝑖𝑛 𝑥. Precisely,
𝑃𝑒𝑟𝑖𝑜𝑑 =
2π
|𝐵|
(since period is positive (a magnitude)). This appears to fly in the face of my claim that the
period of 𝑦 = 𝑠𝑖𝑛 𝑥 is 2π radians since that is how many unique angles there are, because
1
𝐵
4
there are still only [0, 2π[ unique angles for 𝑦 = 𝑠𝑖𝑛(𝐵𝑥). The subtlety comes from the fact
that the input angle into the sine function is 𝐵𝑥, not 𝑥. For this input angle, it is true that
𝑠𝑖𝑛(𝐵𝑥) = 𝑠𝑖𝑛(𝐵𝑥 + 2π)
but since the relationship between 𝑥 and 𝑠𝑖𝑛(𝐵𝑥) is being plotted rather than the relationship
2π
|𝐵|
2π
))
𝐵
between 𝐵𝑥 and 𝑠𝑖𝑛(𝐵𝑥), we have a period of
𝑠𝑖𝑛(𝐵𝑥 + 2π) = 𝑠𝑖𝑛(𝐵𝑥 + 𝐵(
because
= 𝑠𝑖𝑛(𝐵(𝑥 +
2π
𝐵
)) = 𝑠𝑖𝑛(𝐵𝑥)
The varying period of 𝑥 arises as a result of the fact that 2π is how many unique angles there
are, not in opposition to it.
The role of the constant 𝐶 in 𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷
Figure 4: I graphed 𝑓 for different values of 𝐶, ignoring the other constants.
I notice 𝐶 shifts every 𝑥-value on 𝑦 = 𝑠𝑖𝑛 𝑥 by − 𝐶 units because 𝑥-values − 𝐶 units the
𝑥-values of 𝑦 = 𝑠𝑖𝑛 𝑥 must be substituted into 𝑦 = 𝑠𝑖𝑛(𝑥 + 𝐶) to achieve the same 𝑦-values
as 𝑦 = 𝑠𝑖𝑛 𝑥: 𝑦 = 𝑠𝑖𝑛((𝑥 − 𝐶) + 𝐶) = 𝑠𝑖𝑛 𝑥.
𝐶 does not affect the amplitude because it does not affect 𝑦-values.
𝐶 does not affect the period because all 𝑥-values are shifted by the same number of units, so
the magnitude of 𝑥-values over which the 𝑦-values of the graph repeat themselves does not
change.
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The role of the constant 𝐷 in 𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷
Figure 5: I graphed 𝑓 for different values of 𝐷, ignoring the other constants.
I notice 𝐷 shifts every 𝑦-value of 𝑦 = 𝑠𝑖𝑛 𝑥 by 𝐷 units due to it being added or subtracted from
𝑠𝑖𝑛 𝑥.
𝐷 does not affect the amplitude because all 𝑦-values are shifted the same number of units, so
the difference between any two 𝑦-values, including the difference between the maximum and
minimum that works to define the amplitude, stays constant.
𝐷 does not affect the period because it does not affect 𝑥-values.
Putting everything together
Even though I examined the effect of each of the constants in isolation, leaving the other
constants in their trivial cases (𝐴 = 1, 𝐵 = 1, 𝐶 = 0, 𝐷 = 0), my reasoning holds even when
all of them are in play non-trivially at the same time in the function
𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷. This is because they all affect 𝑓 in different ways and, as laid
out in this form, do not act upon each other.
Based on this form, I can also deduce that stretches, signified by 𝐴 and 𝐵, are always applied
before shifts, signified by 𝐶 and 𝐷. In the case of vertical transformations, this is clear; however,
it is less obvious for horizontal transformations as the parentheses (𝐵(𝑥 + 𝐶)) seem to suggest
the horizontal shift is to be done first. The reason this is not the case is because a horizontal
6
1
shift followed by a horizontal stretch - ( 𝐵 (𝑥 − 𝐶)) - substituted into 𝑦 = 𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) does
not achieve the same 𝑦-values as 𝑦 = 𝑠𝑖𝑛 𝑥:
1
𝑥
𝑦 = 𝑠𝑖𝑛(𝐵(( 𝐵 (𝑥 − 𝐶)) + 𝐶)) = 𝑠𝑖𝑛(𝑥 − 𝐶 + 𝐶𝐵) ≠ 𝑠𝑖𝑛 𝑥 = 𝑠𝑖𝑛(𝐵(( 𝐵 − 𝐶) + 𝐶)
𝑥
By contrast, the substitution of ( 𝐵 − 𝐶) - stretch then shift - does.
Therefore in summary the constants 𝐴, 𝐵, 𝐶, and 𝐷 transform the function
𝑓(𝑥) = 𝐴(𝑠𝑖𝑛(𝐵(𝑥 + 𝐶)) + 𝐷 from 𝑦 = 𝑠𝑖𝑛 𝑥 in the following way, in the following order:
Horizontal transformations:
1) A horizontal stretch by factor
1
𝐵
2) A horizontal shift by − 𝐶 units
Vertical transformations:
1) A vertical stretch by factor 𝐴
2) A vertical shift by 𝐷 units
I checked these predictions for a variety of functions and their graphs matched, suggesting they
are correct.
Figure 6: Graph of 𝑓(𝑥) =
1
2
𝑠𝑖𝑛(𝑥 −
π
4
) + 2 alongside 𝑦 = 𝑠𝑖𝑛 𝑥. I used the graph of
𝑦 = 𝑠𝑖𝑛 𝑥 to inform my sketch of 𝑓 based on the aforementioned transformations as well as
checking exact points as a result of the transformations. The transformations were
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1) A horizontal shift by +
π
4
units
2) A vertical stretch by factor
1
2
3) A vertical shift by + 2 units
Figure 7: My graph agreed with what I graphed on my graphing calculator.
Figure 8: Graph of 𝑔(𝑥) = 2𝑠𝑖𝑛(3(𝑥 +
π
3
)) + 1 alongside 𝑦 = 𝑠𝑖𝑛 𝑥 and 𝑦 = 2𝑠𝑖𝑛(3𝑥) + 1.
Here, due to the number of different transformations, I graphed 𝑦 = 2𝑠𝑖𝑛(3𝑥) + 1 as a
dashed line before applying the horizontal shift as an intermediary step. The transformations
were
1) A horizontal stretch by factor
1
3
8
2) A horizontal shift by −
π
3
units
3) A vertical stretch by factor 2
4) A vertical shift by + 1 units
Figure 9: My graph also matched that on my calculator.
1
Figure 10: Graph of ℎ(𝑥) =− 𝑠𝑖𝑛( 2 𝑥) − 3 alongside 𝑦 = 𝑠𝑖𝑛 𝑥. The transformations were
1) A horizontal stretch by factor 2
2) A vertical stretch by factor − 1
3) A vertical shift by − 3 units
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Figure 11: Again, my drawing agreed with what I graphed on my calculator.