Sinusoidal Equations from Graphs

Going backwards, from a wave to its equation, is what lets you build models of real data. You read four things off the graph (amplitude, axis, period, and a starting point) and the equation almost writes itself. All angles are in degrees.

Key ideas

Four steps

For $y = a\sin\big(k(x - d)\big) + c$ or $y = a\cos\big(k(x - d)\big) + c$:

1. Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$
2. Axis: $c = \dfrac{\text{max} + \text{min}}{2}$
3. Period, then $k = \dfrac{360^\circ}{\text{period}}$. Measure the period between two matching points, like consecutive maximums.
4. Phase shift $d$:
   • For cosine, choose $d$ at a maximum: $y = a\cos x$ starts at its peak.
   • For sine, choose $d$ where the graph crosses the axis going up: $y = a\sin x$ starts on its axis, rising.

More than one right answer

The same graph has many correct equations. You can use sine or cosine, choose a different maximum (any $d$ plus or minus a whole period works), or use a negative $a$ and start from a minimum. Check any equation by substituting a known point.

Worked examples

Example 1: From a graph

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

Solution.

1. $a = \dfrac{5 - (-1)}{2} = 3$
2. $c = \dfrac{5 + (-1)}{2} = 2$
3. The maximums are at $60^\circ$ and $240^\circ$, so the period is $180^\circ$ and $k = \dfrac{360^\circ}{180^\circ} = 2$.
4. Cosine: there’s a maximum at $x = 60^\circ$, so $d = 60^\circ$:

$$y = 3\cos\big(2(x - 60^\circ)\big) + 2$$

Sine: the graph crosses the axis going up halfway between the minimum at $-30^\circ$ (one period before $150^\circ$) and the maximum at $60^\circ$. That’s a quarter period before the maximum: $60^\circ - 45^\circ = 15^\circ$:

$$y = 3\sin\big(2(x - 15^\circ)\big) + 2$$

Check both at $x = 60^\circ$: $3\cos 0^\circ + 2 = 5$ ✓ and $3\sin 90^\circ + 2 = 5$ ✓.

Example 2: From a description

A sinusoidal function has a maximum value of $10$ at $x = 30^\circ$, a minimum value of $4$, and a period of $120^\circ$. Write a cosine equation.

Solution. $a = \tfrac{10 - 4}{2} = 3$, $c = \tfrac{10 + 4}{2} = 7$, $k = \tfrac{360^\circ}{120^\circ} = 3$, and the maximum gives $d = 30^\circ$:

$$y = 3\cos\big(3(x - 30^\circ)\big) + 7$$

Check: half a period after the maximum, at $x = 90^\circ$, $y = 3\cos 180^\circ + 7 = 4$, the minimum. ✓

Example 3: From a table

Write an equation for this data.

$x$	$0^\circ$	$45^\circ$	$90^\circ$	$135^\circ$	$180^\circ$
$y$	$1$	$4$	$1$	$-2$	$1$

Solution. Maximum $4$ at $45^\circ$, minimum $-2$ at $135^\circ$, and the pattern repeats after $180^\circ$.

$a = 3$, $c = 1$, period $180^\circ$ so $k = 2$. At $x = 0^\circ$, $y = 1$ is on the axis and rising, so a sine function with $d = 0$ fits:

$$y = 3\sin(2x) + 1$$

Check: $x = 45^\circ$ gives $3\sin 90^\circ + 1 = 4$. ✓

Example 4: Starting from a minimum

Write a second equation for the graph in Example 1 using a cosine with a negative $a$.

Solution. $y = -3\cos x$ starts at its minimum. The graph has a minimum at $150^\circ$, so use $d = 150^\circ$:

$$y = -3\cos\big(2(x - 150^\circ)\big) + 2$$

Check at $x = 60^\circ$: $-3\cos(-180^\circ) + 2 = -3(-1) + 2 = 5$. ✓

Common mistakes

Using the full height for $a$. The amplitude is half of (maximum − minimum).

Using the period as $k$. $k = \dfrac{360^\circ}{\text{period}}$, so a period of $180^\circ$ gives $k = 2$.

Choosing the wrong starting point. For cosine, $d$ is at a maximum (or a minimum if $a \lt 0$). For sine, $d$ is where the graph crosses its axis going up, not where it crosses the $x$-axis.

Not checking. Substitute a maximum or minimum into your equation. If it doesn’t give the right $y$, revisit $d$.

Practice

1. (Warm-up) A sinusoidal graph has a maximum of $8$ and a minimum of $2$. Find $a$ (positive) and $c$.

Solution

$a = 3$, $c = 5$.

2. (Warm-up) A sinusoidal graph has a period of $90^\circ$. Find $k$.

Solution

$k = \tfrac{360^\circ}{90^\circ} = 4$.

3. (Warm-up) A graph has a maximum at $x = 20^\circ$. What value of $d$ would you use in a cosine equation with $a \gt 0$?

Solution

$d = 20^\circ$.

4. (Core) A sinusoidal function has a maximum of $6$ at $x = 0^\circ$ and its next minimum, $-2$, at $x = 180^\circ$. Write its equation.

Solution

$a = 4$, $c = 2$. Maximum to minimum is half a period, so the period is $360^\circ$ and $k = 1$. Cosine with $d = 0$:

$$y = 4\cos x + 2$$

5. (Core) A sinusoidal function has a maximum of $3$, a minimum of $-3$, and a period of $720^\circ$, and it passes through the origin going up. Write its equation.

Solution

$a = 3$, $c = 0$, $k = \tfrac{360^\circ}{720^\circ} = 0.5$, and it starts on its axis rising, so sine with $d = 0$:

$$y = 3\sin(0.5x)$$

6. (Core) Write a cosine equation for this data.

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$
$y$	$5$	$2$	$-1$	$2$	$5$

Solution

Maximum $5$ at $0^\circ$ and $120^\circ$, so the period is $120^\circ$ and $k = 3$. Minimum $-1$, so $a = 3$ and $c = 2$:

$$y = 3\cos(3x) + 2$$

7. (Core) Write a sine equation for the data in Question 6.

Solution

The graph crosses its axis $y = 2$ going up at $x = 90^\circ$ (between the minimum at $60^\circ$ and the maximum at $120^\circ$). So $d = 90^\circ$:

$$y = 3\sin\big(3(x - 90^\circ)\big) + 2$$

Check $x = 0^\circ$: $3\sin(-270^\circ) + 2 = 3(1) + 2 = 5$. ✓

8. (Challenge) Explain why $y = 3\cos\big(2(x - 240^\circ)\big) + 2$ is also a correct equation for the graph in Example 1.

Solution

The graph has a maximum at $240^\circ$ too, so it’s equally valid to start the cosine there. Also, $240^\circ = 60^\circ + 180^\circ$: shifting by a whole period gives the same graph.

9. (Challenge) A sinusoidal graph has a minimum of $-5$ at $x = 30^\circ$ and the next maximum, $1$, at $x = 90^\circ$. Write a cosine equation with $a \lt 0$, and a cosine equation with $a \gt 0$.

Solution

$|a| = 3$, $c = -2$. Minimum to maximum is half a period, so the period is $120^\circ$ and $k = 3$.

With $a \lt 0$, start at the minimum: $y = -3\cos\big(3(x - 30^\circ)\big) - 2$.

With $a \gt 0$, start at the maximum: $y = 3\cos\big(3(x - 90^\circ)\big) - 2$.