Sinusoidal Functions in Radians

In Grade 11 you graphed $y = \sin x$ and $y = \cos x$ with $x$ in degrees and transformed them. Now the $x$-axis is in radians. The shapes are exactly the same; only the numbers on the $x$-axis change, with $2\pi$ in place of $360^\circ$. Radians are what you’ll use in calculus and science, so this is the version of these graphs you’ll see from now on.

Key ideas

The graphs of sin x and cos x in radians

Plot $y = \sin x$ and $y = \cos x$ using the unit circle values at the quadrantal angles:

$x$	$0$	$\tfrac{\pi}{2}$	$\pi$	$\tfrac{3\pi}{2}$	$2\pi$
$\sin x$	$0$	$1$	$0$	$-1$	$0$
$\cos x$	$1$	$0$	$-1$	$0$	$1$

These are the five key points of one cycle.

One cycle of $y = \sin x$ and $y = \cos x$, with the key points every $\tfrac{\pi}{2}$.

Property	$y = \sin x$	$y = \cos x$
Period	$2\pi$	$2\pi$
Amplitude	$1$	$1$
Axis	$y = 0$	$y = 0$
Domain	$\{x \in \mathbb{R}\}$	$\{x \in \mathbb{R}\}$
Range	$\{y \in \mathbb{R} \mid -1 \le y \le 1\}$	$\{y \in \mathbb{R} \mid -1 \le y \le 1\}$
Zeros	$x = 0, \pm\pi, \pm 2\pi, \ldots$	$x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \ldots$
Maximum $1$ at	$x = \tfrac{\pi}{2} + 2n\pi$	$x = 2n\pi$

Here $n$ is any integer. Notice that the cosine graph is the sine graph shifted $\tfrac{\pi}{2}$ to the left.

The transformed function

Every sinusoidal function can be written as

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

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 $= \dfrac{2\pi}{\lvert k \rvert}$
$d$	horizontal translation	phase shift ($d \gt 0$ right, $d \lt 0$ left)
$c$	vertical translation	axis $y = c$

The only change from Grade 11 is the period: $\dfrac{2\pi}{\lvert k \rvert}$ instead of $\dfrac{360^\circ}{\lvert k \rvert}$. Going the other way, $k = \dfrac{2\pi}{\text{period}}$. The maximum is $c + \lvert a \rvert$ and the minimum is $c - \lvert a \rvert$.

Sketching with key points

Apply the mapping rule from combined transformations to the five key points:

$$(x, y) \to \left(\frac{x}{k} + d,\ ay + c\right)$$

As always, factor out $k$ first. For example, $\sin\left(2x - \tfrac{\pi}{2}\right) = \sin\left(2\left(x - \tfrac{\pi}{4}\right)\right)$, so the phase shift is $\tfrac{\pi}{4}$, not $\tfrac{\pi}{2}$. Once you have one cycle, repeat it by adding or subtracting the period.

Writing an equation from a graph or properties

1. Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$
2. Axis: $c = \dfrac{\text{max} + \text{min}}{2}$
3. Period, then $k = \dfrac{2\pi}{\text{period}}$
4. Phase shift $d$: for cosine, the $x$-value of a maximum; for sine, the $x$-value where the graph crosses the axis going up. (With a negative $a$, cosine can start at a minimum instead.)

Every graph has many correct equations. Check yours by substituting a known point.

Worked examples

Example 1: Reading the properties

State the amplitude, period, phase shift, axis, maximum, minimum, and range of $y = 3\cos\left(2x - \dfrac{\pi}{3}\right) + 1$.

Solution. First factor out $k = 2$ from the bracket:

$$2x - \frac{\pi}{3} = 2\left(x - \frac{\pi}{6}\right) \quad\Rightarrow\quad y = 3\cos\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$$

So $a = 3$, $k = 2$, $d = \tfrac{\pi}{6}$, $c = 1$:

• amplitude $3$
• period $\dfrac{2\pi}{2} = \pi$
• phase shift $\dfrac{\pi}{6}$ to the right
• axis $y = 1$
• maximum $1 + 3 = 4$, minimum $1 - 3 = -2$
• range $\{y \in \mathbb{R} \mid -2 \le y \le 4\}$

Example 2: Sketching from key points

Sketch $y = 2\sin\left(2\left(x - \dfrac{\pi}{4}\right)\right) + 1$ for $0 \le x \le 2\pi$.

Solution. Here $a = 2$, $k = 2$, $d = \tfrac{\pi}{4}$, $c = 1$, so the mapping is $(x, y) \to \left(\dfrac{x}{2} + \dfrac{\pi}{4},\ 2y + 1\right)$. Map the key points of $y = \sin x$:

$y = \sin x$	$(0, 0)$	$\left(\tfrac{\pi}{2}, 1\right)$	$(\pi, 0)$	$\left(\tfrac{3\pi}{2}, -1\right)$	$(2\pi, 0)$
image	$\left(\tfrac{\pi}{4}, 1\right)$	$\left(\tfrac{\pi}{2}, 3\right)$	$\left(\tfrac{3\pi}{4}, 1\right)$	$(\pi, -1)$	$\left(\tfrac{5\pi}{4}, 1\right)$

For example, $\left(\tfrac{\pi}{2}, 1\right) \to \left(\tfrac{\pi}{4} + \tfrac{\pi}{4},\ 2(1) + 1\right) = \left(\tfrac{\pi}{2}, 3\right)$.

The period is $\pi$, so add $\pi$ to repeat: there’s another maximum at $\left(\tfrac{3\pi}{2}, 3\right)$ and another minimum at $(2\pi, -1)$. Going back one quarter period from $\left(\tfrac{\pi}{4}, 1\right)$ gives a minimum at $(0, -1)$.

$y = 2\sin\left(2\left(x - \tfrac{\pi}{4}\right)\right) + 1$ (solid) and $y = \sin x$ (dashed). The mapped key points are marked.

Check: at $x = \tfrac{\pi}{2}$, $y = 2\sin\left(2 \cdot \tfrac{\pi}{4}\right) + 1 = 2\sin\tfrac{\pi}{2} + 1 = 3$. ✓

Example 3: An equation from a graph, two ways

Write a cosine equation and a sine equation for this graph.

Solution. The maximum is $4$ and the minimum is $-2$:

$$a = \frac{4 - (-2)}{2} = 3 \qquad\qquad c = \frac{4 + (-2)}{2} = 1$$

From one minimum, $(0, -2)$, to the next, $(\pi, -2)$, is one cycle, so the period is $\pi$ and $k = \dfrac{2\pi}{\pi} = 2$.

Cosine: a maximum is at $x = \tfrac{\pi}{2}$, so $d = \tfrac{\pi}{2}$:

$$y = 3\cos\left(2\left(x - \frac{\pi}{2}\right)\right) + 1$$

Sine: the graph crosses the axis going up halfway between the minimum at $0$ and the maximum at $\tfrac{\pi}{2}$, at $x = \tfrac{\pi}{4}$. So $d = \tfrac{\pi}{4}$:

$$y = 3\sin\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$$

Check both at $(0, -2)$: $3\cos(-\pi) + 1 = -3 + 1 = -2$ ✓ and $3\sin\left(-\tfrac{\pi}{2}\right) + 1 = -3 + 1 = -2$ ✓.

A third option: the graph starts at a minimum at $x = 0$, so $y = -3\cos(2x) + 1$ also works.

Example 4: An equation from properties

A sinusoidal function has an amplitude of $5$, a period of $4\pi$, and a minimum at $(\pi, -3)$. Write its equation as a cosine function and as a sine function.

Solution. $a = 5$ and $k = \dfrac{2\pi}{4\pi} = \dfrac{1}{2}$. The minimum is $c - 5 = -3$, so $c = 2$ (the maximum is $7$).

Cosine: a maximum is half a period ($2\pi$) away from the minimum, at $x = \pi + 2\pi = 3\pi$:

$$y = 5\cos\left(\frac{1}{2}(x - 3\pi)\right) + 2$$

Sine: the graph crosses the axis going up a quarter period ($\pi$) after the minimum, at $x = \pi + \pi = 2\pi$:

$$y = 5\sin\left(\frac{1}{2}(x - 2\pi)\right) + 2$$

Check both at $x = \pi$: $5\cos(-\pi) + 2 = -3$ ✓ and $5\sin\left(-\tfrac{\pi}{2}\right) + 2 = -3$ ✓.

(You could also start the cosine at the minimum with a negative $a$: $y = -5\cos\left(\tfrac{1}{2}(x - \pi)\right) + 2$.)

Common mistakes

Using 360 in the period. In radians, the period is $\dfrac{2\pi}{\lvert k \rvert}$. For $y = \sin 4x$ it’s $\tfrac{2\pi}{4} = \tfrac{\pi}{2}$, not $90$.

Not factoring out k. In $y = \cos\left(3x + \tfrac{\pi}{2}\right)$, the phase shift is not $\tfrac{\pi}{2}$. Factor: $3\left(x + \tfrac{\pi}{6}\right)$, so it’s $\tfrac{\pi}{6}$ to the left.

Getting the direction of the phase shift backwards. $(x - d)$ shifts right by $d$; $\left(x + \tfrac{\pi}{6}\right) = \left(x - \left(-\tfrac{\pi}{6}\right)\right)$ shifts left.

Mixing degrees and radians. An equation like $y = \sin\left(2(x - 45)\right)$ on a radian graph means a shift of $45$ radians. Write $\tfrac{\pi}{4}$.

Using a maximum for the sine phase shift. Sine starts on its axis going up, not at a maximum. Only cosine starts at a maximum.

Checking with the calculator in degree mode. When you substitute a point to check your equation, use radian mode.

Practice

1. (Warm-up) State the amplitude, period, phase shift, and axis of $y = 5\sin\left(4\left(x + \dfrac{\pi}{3}\right)\right) - 2$.

Solution

Amplitude $5$, period $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$, phase shift $\dfrac{\pi}{3}$ to the left, axis $y = -2$.

2. (Warm-up) Find $k$ (positive) for a period of (a) $\pi$ (b) $6\pi$ (c) $3$.

Solution

$k = \dfrac{2\pi}{\text{period}}$:

(a) $k = \dfrac{2\pi}{\pi} = 2$ (b) $k = \dfrac{2\pi}{6\pi} = \dfrac{1}{3}$ (c) $k = \dfrac{2\pi}{3}$

3. (Warm-up) List all the zeros of $y = \sin x$ for $-2\pi \le x \le 2\pi$, and all the maximum points of $y = \cos x$ in the same interval.

Solution

Zeros of $\sin x$: $x = -2\pi, -\pi, 0, \pi, 2\pi$.

Maximum points of $\cos x$: $(-2\pi, 1)$, $(0, 1)$, $(2\pi, 1)$.

4. (Core) State the amplitude, period, phase shift, axis, maximum, and minimum of $y = -2\cos\left(\dfrac{1}{2}x + \dfrac{\pi}{4}\right) + 3$. What does the negative sign do?

Solution

Factor out $\tfrac{1}{2}$: $\tfrac{1}{2}x + \tfrac{\pi}{4} = \tfrac{1}{2}\left(x + \tfrac{\pi}{2}\right)$. So $a = -2$, $k = \tfrac{1}{2}$, $d = -\tfrac{\pi}{2}$, $c = 3$.

Amplitude $2$, period $\dfrac{2\pi}{1/2} = 4\pi$, phase shift $\dfrac{\pi}{2}$ to the left, axis $y = 3$, maximum $5$, minimum $1$.

The negative sign reflects the graph in its axis, so this cosine starts its cycle (at $x = -\tfrac{\pi}{2}$) at a minimum instead of a maximum.

5. (Core) Use the mapping rule to find the images of the five key points of $y = \cos x$ (from $0$ to $2\pi$) under $y = 3\cos\left(2\left(x + \dfrac{\pi}{6}\right)\right) - 1$.

Solution

The mapping is $(x, y) \to \left(\dfrac{x}{2} - \dfrac{\pi}{6},\ 3y - 1\right)$:

• $(0, 1) \to \left(-\tfrac{\pi}{6}, 2\right)$
• $\left(\tfrac{\pi}{2}, 0\right) \to \left(\tfrac{\pi}{4} - \tfrac{\pi}{6}, -1\right) = \left(\tfrac{\pi}{12}, -1\right)$
• $(\pi, -1) \to \left(\tfrac{\pi}{2} - \tfrac{\pi}{6}, -4\right) = \left(\tfrac{\pi}{3}, -4\right)$
• $\left(\tfrac{3\pi}{2}, 0\right) \to \left(\tfrac{3\pi}{4} - \tfrac{\pi}{6}, -1\right) = \left(\tfrac{7\pi}{12}, -1\right)$
• $(2\pi, 1) \to \left(\pi - \tfrac{\pi}{6}, 2\right) = \left(\tfrac{5\pi}{6}, 2\right)$

Check one: at $x = \tfrac{\pi}{3}$, $y = 3\cos\left(2 \cdot \tfrac{\pi}{2}\right) - 1 = 3\cos\pi - 1 = -4$. ✓

6. (Core) A sinusoidal function has a maximum value of $5$, a minimum value of $-1$, a period of $\dfrac{2\pi}{3}$, and a maximum at $x = \dfrac{\pi}{6}$. Write its equation in two different ways.

Solution

$a = \dfrac{5 - (-1)}{2} = 3$, $c = \dfrac{5 + (-1)}{2} = 2$, and $k = \dfrac{2\pi}{2\pi/3} = 3$.

Cosine, with $d$ at the maximum: $y = 3\cos\left(3\left(x - \dfrac{\pi}{6}\right)\right) + 2$.

Sine: the graph crosses the axis going up a quarter period ($\tfrac{1}{4} \cdot \tfrac{2\pi}{3} = \tfrac{\pi}{6}$) before the maximum, at $x = 0$. So $y = 3\sin(3x) + 2$.

Check the sine version at the maximum: $3\sin\left(3 \cdot \tfrac{\pi}{6}\right) + 2 = 3\sin\tfrac{\pi}{2} + 2 = 5$. ✓

7. (Core) A sinusoidal graph has a maximum at $\left(\dfrac{\pi}{3}, 6\right)$, and the next minimum is at $\left(\dfrac{4\pi}{3}, -2\right)$. Write a cosine equation and a sine equation for it.

Solution

$a = \dfrac{6 - (-2)}{2} = 4$ and $c = \dfrac{6 + (-2)}{2} = 2$. Maximum to the next minimum is half a period, $\tfrac{4\pi}{3} - \tfrac{\pi}{3} = \pi$, so the period is $2\pi$ and $k = 1$.

Cosine: $y = 4\cos\left(x - \dfrac{\pi}{3}\right) + 2$.

Sine: the upward axis crossing is a quarter period ($\tfrac{\pi}{2}$) before the maximum, at $\tfrac{\pi}{3} - \tfrac{\pi}{2} = -\tfrac{\pi}{6}$. So $y = 4\sin\left(x + \dfrac{\pi}{6}\right) + 2$.

Check the sine version at $x = \tfrac{4\pi}{3}$: $4\sin\tfrac{3\pi}{2} + 2 = -4 + 2 = -2$. ✓

8. (Challenge) A sinusoidal function has a maximum value of $7$ at $x = \dfrac{\pi}{8}$, and the closest minimum to its right is $-1$, at $x = \dfrac{3\pi}{8}$. Write its equation, and find the exact value of $y$ when $x = \dfrac{\pi}{24}$.

Solution

$a = 4$ and $c = 3$. Half a period is $\tfrac{3\pi}{8} - \tfrac{\pi}{8} = \tfrac{\pi}{4}$, so the period is $\tfrac{\pi}{2}$ and $k = \dfrac{2\pi}{\pi/2} = 4$. With cosine and the maximum at $\tfrac{\pi}{8}$:

$$y = 4\cos\left(4\left(x - \frac{\pi}{8}\right)\right) + 3$$

At $x = \tfrac{\pi}{24}$:

$$\begin{aligned} y &= 4\cos\left(4\left(\frac{\pi}{24} - \frac{3\pi}{24}\right)\right) + 3 \\ &= 4\cos\left(-\frac{\pi}{3}\right) + 3 \\ &= 4\left(\frac{1}{2}\right) + 3 = 5 \end{aligned}$$

9. (Challenge) Explain why $y = 2\sin(3x - \pi) + 1$ and $y = -2\sin(3x) + 1$ have the same graph.

Solution

Factor the first one: $3x - \pi = 3\left(x - \tfrac{\pi}{3}\right)$, so $y = 2\sin\left(3\left(x - \tfrac{\pi}{3}\right)\right) + 1$. Its period is $\tfrac{2\pi}{3}$, so the phase shift of $\tfrac{\pi}{3}$ is exactly half a period.

Shifting a sine graph by half a period moves each maximum to where a minimum was, and each minimum to where a maximum was, while the axis crossings stay on the axis. That’s the same as reflecting the graph in its axis, which is what the negative $a$ does in $y = -2\sin(3x) + 1$. Both graphs have amplitude $2$, period $\tfrac{2\pi}{3}$, and axis $y = 1$, and both pass through $(0, 1)$ going down.

Check at $x = \tfrac{\pi}{6}$: $2\sin\left(\tfrac{\pi}{2} - \pi\right) + 1 = 2(-1) + 1 = -1$, and $-2\sin\tfrac{\pi}{2} + 1 = -1$. ✓