What does it mean if b is negative?

A negative b reflects the graph horizontally, but since the period formula uses |b|, the magnitude of the period remains unchanged. For clarity, you can rewrite \sin(-bx) as -\sin(bx) to separate the reflection from the frequency.

Why is the period formula \frac{2\pi}{|b|} and not something else?

The parent function \sin(x) completes one full cycle over an interval of 2\pi because \sin(x + 2\pi) = \sin(x). When you compose sine with the argument bx, the period becomes \frac{2\pi}{|b|} because the input is scaled by a factor of b.

How do I find the maximum and minimum values of the function?

Since \sin(\theta) ranges from -1 to 1, the maximum of y = a\sin(b(x - h)) + k is k + |a| and the minimum is k - |a|. The vertical shift k moves the entire range up or down.

Can I use this formula with functions other than sine?

Yes, the same amplitude and period formulas work for cosine: y = a\cos(b(x - h)) + k. For tangent, the period formula is different—\frac{\pi}{|b|} instead of \frac{2\pi}{|b|}—because tangent has period \pi, not 2\pi.

Amplitude and Period

Amplitude and Period tell you how tall and wide a sine wave is, used when analyzing or graphing trigonometric functions and oscillating systems.

y = a\sin\big(b(x - h)\big) + k, \quad \text{period} = \frac{2\pi}{|b|}

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What each symbol means

What Amplitude and Period takes

Symbol	Meaning
$a$	The amplitude coefficient; if negative, the wave reflects over the x-axis, but amplitude is always the absolute value $\|a\|$ .
$b$	The frequency factor determining the period; larger $\|b\|$ means the oscillation completes more quickly.
$h$	The horizontal shift (phase shift); positive $h$ shifts the graph right, negative $h$ shifts left.
$k$	The vertical shift; the graph shifts up if $k > 0$ and down if $k < 0$ .
$x$	The independent variable (usually time, angle, or distance); the input to the sine function.
$y$	The dependent variable (height or value of the oscillation); the output of the sine function.

When to use it

When you need to graph a sine or cosine function, or to analyze the behavior of any oscillating pattern.

Level

Usually taught in: Algebra II · Appears on: SAT

Worked examples

1. Find amplitude and period of a basic sine function

Problem

Find the amplitude and period of

y = 3\sin(2x) + 1

$y = 3\sin(2x) + 1$
The given function is in the form $y = a\sin(b(x - h)) + k$ with $h = 0$ and $k = 1$ .
$a = 3, \quad b = 2, \quad h = 0, \quad k = 1$
We identify the amplitude coefficient $a = 3$ and the frequency coefficient $b = 2$ .
$\text{Amplitude} = |a| = 3$
The amplitude is the absolute value of $a$ , which is 3.
$\text{Period} = \frac{2\pi}{|b|} = \frac{2\pi}{2} = \pi$
We apply the period formula with $b = 2$ to get $\pi$ .

Answer: $\text{Amplitude} = 3, \quad \text{Period} = \pi$

This basic example shows how to directly apply the formulas for amplitude and period to a function already in standard form. The amplitude of 3 means the wave oscillates 3 units above and below the center line of $y = 1$ , and the period of $\pi$ means each complete cycle spans $\pi$ units along the x-axis.

2. Find amplitude and period with a negative coefficient and fraction

Problem

Find the amplitude and period of

y = -\frac{2}{3}\sin(3x) + 4

$y = -\frac{2}{3}\sin(3x) + 4$
The given function is in the form $y = a\sin(bx) + k$ , with $h = 0$ .
$a = -\frac{2}{3}, \quad b = 3, \quad k = 4$
We identify each coefficient by comparing to the standard form.
$\text{Amplitude} = |a| = \left|-\frac{2}{3}\right| = \frac{2}{3}$
Even though $a$ is negative, the amplitude is always positive; we take the absolute value.
$\text{Period} = \frac{2\pi}{|b|} = \frac{2\pi}{3}$
We apply the period formula with $b = 3$ .

Answer: $\text{Amplitude} = \frac{2}{3}, \quad \text{Period} = \frac{2\pi}{3}$

This example highlights that a negative coefficient $a$ does not make the amplitude negative—it only flips the graph. The amplitude remains the positive value $\frac{2}{3}$ , and the frequency factor $b = 3$ compresses the period to one-third of the standard $2\pi$ .

3. Find amplitude and period for a two-leg road trip elevation model

Problem

On a road trip over a mountain pass, the elevation along the first leg can be modeled by

h(x) = 150\sin(0.4x) + 1800

, where

h

is elevation in feet and

x

is horizontal distance in miles. Find the elevation amplitude and the horizontal distance between consecutive peaks.

$h(x) = 150\sin(0.4x) + 1800$
This elevation model is in the form $y = a\sin(bx) + k$ , with $h = 0$ .
$a = 150, \quad b = 0.4 = \frac{2}{5}, \quad k = 1800$
We identify the parameters; note that $0.4 = \frac{2}{5}$ , which will simplify the period calculation.
$\text{Amplitude} = |a| = 150 \text{ feet}$
The amplitude tells us how far the elevation swings above and below the average elevation of 1800 feet.
$\text{Period} = \frac{2\pi}{|b|} = \frac{2\pi}{0.4} = \frac{2\pi}{\frac{2}{5}} = 2\pi \cdot \frac{5}{2} = 5\pi \text{ miles}$
We apply the period formula; dividing by 0.4 is equivalent to multiplying by its reciprocal $\frac{5}{2}$ .
$5\pi \approx 15.7 \text{ miles}$
Numerically, one complete cycle of the terrain pattern repeats approximately every 15.7 miles along the road.

Answer: $\text{Amplitude} = 150 \text{ feet}; \quad \text{distance between consecutive peaks} = 5\pi \approx 15.7 \text{ miles}$

This real-world example shows how amplitude and period describe physical oscillations. The amplitude of 150 feet represents the maximum elevation change from the baseline of 1800 feet (ranging from 1650 to 1950 feet), while the period of $5\pi$ miles tells us how far apart successive mountain peaks are along the road.

Common mistakes

Where Amplitude and Period usually goes wrong

The mistake	Why it is wrong	The fix
Writing amplitude as $a$ when $a$ is negative, such as 'amplitude = -2' for $a = -2$ .	Amplitude must be non-negative; a negative value of $a$ indicates a reflection over the x-axis, but does not make amplitude negative.	Always compute amplitude as the absolute value: $\text{amplitude} = \|a\|$ .
Using the inverted formula $\text{period} = \frac{\|b\|}{2\pi}$ instead of the correct formula.	This formula gives the frequency (cycles per unit) instead of the period (unit length per cycle); it is the reciprocal of what we need.	Use the correct formula: $\text{period} = \frac{2\pi}{\|b\|}$ .
Ignoring the factor $b$ and claiming the period is always $2\pi$ .	The value $2\pi$ is the period only when $b = 1$ ; any other value of $b$ compresses or stretches the oscillation.	Always divide $2\pi$ by the absolute value of $b$ : $\text{period} = \frac{2\pi}{\|b\|}$ .

Tips and when to use something else

The amplitude is always non-negative; if $a < 0$ , the graph flips upside down, but $|a|$ remains the amplitude.
Larger $|b|$ means tighter oscillations—compare $\sin(x)$ with period $2\pi$ to $\sin(10x)$ with period $\frac{\pi}{5}$ to see the effect.
To find the maximum value of the entire function, use $k + |a|$ ; the minimum is $k - |a|$ .
If you encounter a tangent function like $y = a\tan(b(x - h)) + k$ , use the period formula $\frac{\pi}{|b|}$ instead, since tangent has period $\pi$ rather than $2\pi$ .

Related formulas

Sine Cosine TangentFind missing sides or angles in right triangles using sine, cosine, and tangent ratios—essential for SAT and ACT geometry.Unit CircleThe unit circle is a fundamental tool that shows how angles map to sine and cosine coordinates: the point (cos θ, sin θ) always lies on x² + y² = 1.Solving Trigonometric EquationsFind all angles that produce a specific sine value by using inverse sine and the periodic nature of trigonometric functions.

Frequently asked questions

What does it mean if b is negative?: A negative $b$ reflects the graph horizontally, but since the period formula uses $|b|$ , the magnitude of the period remains unchanged. For clarity, you can rewrite $\sin(-bx)$ as $-\sin(bx)$ to separate the reflection from the frequency.
Why is the period formula $\frac{2\pi}{|b|}$ and not something else?: The parent function $\sin(x)$ completes one full cycle over an interval of $2\pi$ because $\sin(x + 2\pi) = \sin(x)$ . When you compose sine with the argument $bx$ , the period becomes $\frac{2\pi}{|b|}$ because the input is scaled by a factor of $b$ .
How do I find the maximum and minimum values of the function?: Since $\sin(\theta)$ ranges from $-1$ to $1$ , the maximum of $y = a\sin(b(x - h)) + k$ is $k + |a|$ and the minimum is $k - |a|$ . The vertical shift $k$ moves the entire range up or down.
Can I use this formula with functions other than sine?: Yes, the same amplitude and period formulas work for cosine: $y = a\cos(b(x - h)) + k$ . For tangent, the period formula is different— $\frac{\pi}{|b|}$ instead of $\frac{2\pi}{|b|}$ —because tangent has period $\pi$ , not $2\pi$ .

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Reviewed 2026-09-18