Transformations of Sinusoidal Functions

Real periodic patterns rarely have amplitude $1$ and period $360^\circ$. By stretching and shifting $y = \sin x$ and $y = \cos x$, using the same transformations as other functions, you can match any wave. In this setting, the parameters get special names: amplitude, period, phase shift, and axis. All angles are in degrees.

Key ideas

The transformed sinusoidal function

$$y = a\sin\big(k(x - d)\big) + c \qquad\text{or}\qquad y = a\cos\big(k(x - d)\big) + c$$

Parameter	Effect	Name
$a$	vertical stretch by $\lvert a \rvert$; reflection in the axis if $a \lt 0$	amplitude $= \lvert a \rvert$
$k$	horizontal stretch or compression by $\tfrac{1}{\lvert k \rvert}$	period $= \dfrac{360^\circ}{\lvert k \rvert}$
$d$	horizontal translation	phase shift ($d \gt 0$ right, $d \lt 0$ left)
$c$	vertical translation	axis $y = c$

From these:

$$\text{maximum} = c + \lvert a \rvert, \qquad \text{minimum} = c - \lvert a \rvert$$

and the range is $\{y \in \mathbb{R} \mid c - \lvert a \rvert \le y \le c + \lvert a \rvert\}$.

Sketching with key points

Apply the mapping rule $(x, y) \to \left(\dfrac{x}{k} + d,\ ay + c\right)$ to the five key points of one cycle:

• $y = \sin x$: $(0^\circ, 0)$, $(90^\circ, 1)$, $(180^\circ, 0)$, $(270^\circ, -1)$, $(360^\circ, 0)$
• $y = \cos x$: $(0^\circ, 1)$, $(90^\circ, 0)$, $(180^\circ, -1)$, $(270^\circ, 0)$, $(360^\circ, 1)$

As always, factor out $k$ first: $\sin(2x - 60^\circ) = \sin\big(2(x - 30^\circ)\big)$.

Worked examples

Example 1: Reading the properties

State the amplitude, period, phase shift, axis, maximum, minimum, and range of $y = 3\cos\big(2(x + 45^\circ)\big) - 1$.

Solution. $a = 3$, $k = 2$, $d = -45^\circ$, $c = -1$.

• amplitude $3$
• period $\dfrac{360^\circ}{2} = 180^\circ$
• phase shift $45^\circ$ to the left
• axis $y = -1$
• maximum $-1 + 3 = 2$, minimum $-1 - 3 = -4$
• range $\{y \in \mathbb{R} \mid -4 \le y \le 2\}$

Example 2: Sketching from key points

Sketch one cycle of $y = 2\sin\big(2(x - 30^\circ)\big) + 1$.

Solution. The mapping rule is $(x, y) \to \left(\tfrac{x}{2} + 30^\circ,\ 2y + 1\right)$:

$y = \sin x$	$(0^\circ, 0)$	$(90^\circ, 1)$	$(180^\circ, 0)$	$(270^\circ, -1)$	$(360^\circ, 0)$
image	$(30^\circ, 1)$	$(75^\circ, 3)$	$(120^\circ, 1)$	$(165^\circ, -1)$	$(210^\circ, 1)$

Amplitude $2$, period $180^\circ$, phase shift $30^\circ$ right, axis $y = 1$.

One cycle runs from $30^\circ$ to $210^\circ$, which is $180^\circ$ long, matching the period.

Example 3: Factoring out k

Find the period and phase shift of $y = \sin(3x - 90^\circ)$.

Solution. Factor: $y = \sin\big(3(x - 30^\circ)\big)$. So $k = 3$ and $d = 30^\circ$.

The period is $\dfrac{360^\circ}{3} = 120^\circ$, and the phase shift is $30^\circ$ to the right (not $90^\circ$).

Example 4: A reflection

For $y = -4\sin x + 2$, find the maximum and minimum, and where they occur for $0^\circ \le x \le 360^\circ$.

Solution. Amplitude $4$, axis $y = 2$, so the maximum is $6$ and the minimum is $-2$.

Because $a \lt 0$, the graph is flipped: where $\sin x$ has its minimum ($x = 270^\circ$), this graph has its maximum, $-4(-1) + 2 = 6$. Where $\sin x$ has its maximum ($x = 90^\circ$), this graph has its minimum, $-4(1) + 2 = -2$.

Common mistakes

Using $k$ as the period. The period is $\dfrac{360^\circ}{k}$. For $y = \sin 2x$, the period is $180^\circ$, not $2$.

Not factoring out $k$. In $\sin(3x - 90^\circ)$, the phase shift is $30^\circ$, not $90^\circ$.

Getting the phase shift direction wrong. $\sin(x - 60^\circ)$ shifts right $60^\circ$; $\sin(x + 60^\circ)$ shifts left.

Giving a negative amplitude. The amplitude is $\lvert a \rvert$, always positive. A negative $a$ means a reflection.

Forgetting the axis when finding the max and min. The maximum is $c + \lvert a \rvert$, not just $\lvert a \rvert$.

Practice

1. (Warm-up) State the amplitude and period of $y = 5\sin(4x)$.

Solution

Amplitude $5$, period $\tfrac{360^\circ}{4} = 90^\circ$.

2. (Warm-up) State the phase shift and the axis of $y = \cos(x - 60^\circ) + 3$.

Solution

Phase shift $60^\circ$ right; axis $y = 3$.

3. (Warm-up) Find the maximum and minimum of $y = 2\sin x - 5$.

Solution

Maximum $-5 + 2 = -3$, minimum $-5 - 2 = -7$.

4. (Core) Describe all the properties of $y = -3\sin\big(0.5(x + 90^\circ)\big) + 2$: amplitude, period, phase shift, axis, range, and any reflection.

Solution

Amplitude $3$, reflected in the axis (since $a \lt 0$). Period $\tfrac{360^\circ}{0.5} = 720^\circ$. Phase shift $90^\circ$ left. Axis $y = 2$. Range $\{y \in \mathbb{R} \mid -1 \le y \le 5\}$.

5. (Core) Map the five key points of $y = \cos x$ to find one cycle of $y = 4\cos(x - 45^\circ) - 1$.

Solution

The rule is $(x, y) \to (x + 45^\circ,\ 4y - 1)$:

$$(45^\circ, 3), \quad (135^\circ, -1), \quad (225^\circ, -5), \quad (315^\circ, -1), \quad (405^\circ, 3)$$

6. (Core) Find the period and phase shift of $y = \sin(2x + 60^\circ)$.

Solution

Factor: $\sin\big(2(x + 30^\circ)\big)$. Period $180^\circ$, phase shift $30^\circ$ left.

7. (Core) Show that $y = \cos(x - 90^\circ)$ has the same graph as $y = \sin x$.

Solution

Shifting the cosine graph $90^\circ$ right moves its key points $(0^\circ, 1), (90^\circ, 0), (180^\circ, -1), (270^\circ, 0)$ to $(90^\circ, 1), (180^\circ, 0), (270^\circ, -1), (360^\circ, 0)$, which are exactly sine’s key points. So the graphs match.

8. (Challenge) Describe, in order, the transformations that take $y = \cos x$ to $y = 2\cos\big(3(x - 20^\circ)\big) - 4$.

Solution

1. Vertical stretch by a factor of $2$ (amplitude $2$).
2. Horizontal compression by a factor of $\tfrac{1}{3}$ (period $120^\circ$).
3. Translation $20^\circ$ right (phase shift).
4. Translation $4$ down (axis $y = -4$).

9. (Challenge) A function $y = a\sin(kx) + c$, with $a \gt 0$ and $k \gt 0$, has a maximum of $7$, a minimum of $-1$, and a period of $120^\circ$. Find $a$, $k$, and $c$.

Solution

$a = \tfrac{7 - (-1)}{2} = 4$, $c = \tfrac{7 + (-1)}{2} = 3$, and $\tfrac{360^\circ}{k} = 120^\circ$ gives $k = 3$.

So $y = 4\sin(3x) + 3$.