Math Core

Unrolling the unit circle onto an $xy$ -plane turns each trig function into a graph you can read at a glance. Once you know the two basic waves and the four transformations that act on them, you can graph any sinusoid, read its equation off a picture, and see where tangent and the reciprocal functions blow up.

Sine and cosine as waves

Walk around the unit circle and record the height of the point as a function of the angle $x$ . That height is $\sin x$ : it starts at $0$ , climbs to $1$ at $\dfrac{\pi}{2}$ , returns to $0$ at $\pi$ , drops to $-1$ at $\dfrac{3\pi}{2}$ and comes back to $0$ at $2\pi$ . Recording the horizontal coordinate instead gives $\cos x$ , which starts at its maximum.

y = sin x passes through the origin; y = cos x starts at its maximum (0, 1).Open in grapher →

Everything from the previous lesson is visible here. Both graphs have domain all reals, range $[-1, 1]$ and period $2\pi$ . Cosine is symmetric about the $y$ -axis (even); sine is symmetric about the origin (odd). And the cosine graph is the sine graph shifted left by $\dfrac{\pi}{2}$ :

\cos x = \sin\left(x + \frac{\pi}{2}\right).

So there is really one wave shape, called a sinusoid, and either function can describe it.

Each period splits into four equal quarters, and at the quarter marks the wave hits five key points: for sine the heights are $0, 1, 0, -1, 0$ (midline, max, midline, min, midline) and for cosine $1, 0, -1, 0, 1$ .

The four transformations

The general sinusoid

y = A\sin\big(B(x - C)\big) + D \qquad \text{or} \qquad y = A\cos\big(B(x - C)\big) + D, \qquad B > 0.

Amplitude $|A|$ : half the distance from max to min. If $A < 0$ the wave is also reflected over its midline.
Period $\dfrac{2\pi}{B}$ : the length of one cycle. The frequency $\dfrac{B}{2\pi}$ is the number of cycles per unit of $x$ .
Phase shift $C$ : the horizontal shift (right if $C > 0$ ).
Midline $y = D$ . The range is $[D - |A|,\ D + |A|]$ .

Why $\dfrac{2\pi}{B}$ ? One cycle of the basic wave happens as its input runs from $0$ to $2\pi$ . Here the input is $B(x - C)$ , which runs from $0$ to $2\pi$ as $x$ runs from $C$ to $C + \dfrac{2\pi}{B}$ . That same calculation tells you exactly where a cycle starts and ends, which is all you need for graphing.

Common mistake

Factor $B$ out before reading the phase shift. In $y = \cos(3x - \pi)$ the shift is not $\pi$ : rewrite $3x - \pi = 3\left(x - \dfrac{\pi}{3}\right)$ , so the shift is $\dfrac{\pi}{3}$ to the right. Equivalently, solve $3x - \pi = 0$ to find where the cycle starts.

Worked example: Graphing from the key points

Graph one cycle of $y = -3\sin\left(2x - \dfrac{\pi}{2}\right) + 1$ .

Solution. Factor: $2x - \dfrac{\pi}{2} = 2\left(x - \dfrac{\pi}{4}\right)$ . So $A = -3$ , $B = 2$ , $C = \dfrac{\pi}{4}$ , $D = 1$ . The period is $\dfrac{2\pi}{2} = \pi$ , so the cycle runs from $\dfrac{\pi}{4}$ to $\dfrac{5\pi}{4}$ in quarter steps of $\dfrac{\pi}{4}$ .

$x$	$\frac{\pi}{4}$	$\frac{\pi}{2}$	$\frac{3\pi}{4}$	$\pi$	$\frac{5\pi}{4}$
basic sine	$0$	$1$	$0$	$-1$	$0$
$\times(-3)$ , then $+1$	$1$	$-2$	$1$	$4$	$1$

Because $A$ is negative, the wave goes down first: from the midline at $\left(\dfrac{\pi}{4}, 1\right)$ to a minimum of $-2$ , back to the midline, up to a maximum of $4$ , and back.

y = -3sin(2x - π/2) + 1 with its midline y = 1 dashed. Range [-2, 4], period π.Open in grapher →

Writing the equation from a graph

Going backward is just reading the four numbers off the picture.

$D = \dfrac{\max + \min}{2}$ and $|A| = \dfrac{\max - \min}{2}$ .
Measure the period (peak to peak, or twice the distance from a peak to the next valley) and set $B = \dfrac{2\pi}{\text{period}}$ .
Pick a convenient starting point. A peak is the start of a cosine cycle (use $A > 0$ ); a valley starts a cosine cycle with $A < 0$ ; a midline crossing going up starts a sine cycle.

Because the wave repeats, there are infinitely many correct equations; any one that matches the graph is fine.

Worked example: From features to an equation

A sinusoid has a maximum at $(1, 9)$ and the next minimum at $(4, 3)$ . Write an equation for it.

Solution. $D = \dfrac{9 + 3}{2} = 6$ and $A = \dfrac{9 - 3}{2} = 3$ . From a max to the next min is half a period, so the period is $2(4 - 1) = 6$ and $B = \dfrac{2\pi}{6} = \dfrac{\pi}{3}$ . A maximum at $x = 1$ starts a cosine cycle, so $C = 1$ :

y = 3\cos\left(\frac{\pi}{3}(x - 1)\right) + 6.

Check: at $x = 4$ , the input is $\dfrac{\pi}{3} \cdot 3 = \pi$ , and $3\cos\pi + 6 = 3$ . The minimum matches.

Tangent and cotangent

Tangent is $\dfrac{\sin x}{\cos x}$ , so it is undefined wherever $\cos x = 0$ : at $x = \dfrac{\pi}{2} + k\pi$ for every integer $k$ . Near those values the denominator is tiny and the graph shoots off to $\pm\infty$ , giving vertical asymptotes. Between asymptotes, tangent rises from $-\infty$ to $\infty$ , passing through $0$ at multiples of $\pi$ . Its period is $\pi$ , and it has no amplitude because it is unbounded.

y = tan x, period π, with asymptotes (dashed) at odd multiples of π/2.Open in grapher →

For $y = A\tan\big(B(x - C)\big) + D$ the period is $\dfrac{\pi}{B}$ (not $\dfrac{2\pi}{B}$ ). To find the asymptotes, set the inside equal to $\pm\dfrac{\pi}{2}$ and repeat every period.

Cotangent, $\dfrac{\cos x}{\sin x}$ , also has period $\pi$ . Its asymptotes are where $\sin x = 0$ (at multiples of $\pi$ ), and each branch decreases.

Secant and cosecant

Secant and cosecant are reciprocals of cosine and sine, so they are easiest to graph by first sketching the wave as a guide. Wherever the guide wave crosses $0$ , the reciprocal has an asymptote. Wherever the wave reaches $\pm 1$ , the reciprocal touches it at the same point. Between, the reciprocal opens away from the axis in U-shapes. The range of $\sec x$ and $\csc x$ is $(-\infty, -1] \cup [1, \infty)$ , and both have period $2\pi$ .

y = sec x over its guide y = cos x (dashed). Each U-shape touches the cosine wave at a peak or valley.Open in grapher →

Worked example: Asymptotes of a transformed tangent

Find the period of $y = 2\tan\left(\dfrac{x}{3}\right)$ and its asymptotes in $(-6\pi, 6\pi)$ .

Solution. $B = \dfrac{1}{3}$ , so the period is $\dfrac{\pi}{1/3} = 3\pi$ . The central branch runs while $-\dfrac{\pi}{2} < \dfrac{x}{3} < \dfrac{\pi}{2}$ , that is $-\dfrac{3\pi}{2} < x < \dfrac{3\pi}{2}$ . The asymptotes are at $x = \pm\dfrac{3\pi}{2}$ plus multiples of $3\pi$ : in the given interval, $x = -\dfrac{9\pi}{2}, -\dfrac{3\pi}{2}, \dfrac{3\pi}{2}, \dfrac{9\pi}{2}$ .

Tip

Quick check for any sinusoid equation with $A > 0$ : plug in the $x$ -value of a peak. The inside of the cosine should be a multiple of $2\pi$ (or the inside of the sine should be $\dfrac{\pi}{2}$ plus a multiple of $2\pi$ ), and the output should be $D + |A|$ .

Practice

Practice 1

What is the amplitude of $y = -4\cos(3x) + 1$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

What is the period of $y = 5\sin\left(\dfrac{\pi x}{3}\right) - 2$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

What is the phase shift of $y = \sin\left(4x + \pi\right)$ ?

Practice 4

Give the minimum and maximum values of $y = 4\sin(2x - 1) - 1$ .

Separate answers with commas, e.g. 2, -5

Practice 5

Which equation matches the graph?

Maximum 1 at x = 0, minimum -3 at x = 2 and x = -2.Open in grapher →

Practice 6

A sinusoid $y = A\cos\big(B(x - C)\big) + D$ with $B > 0$ has a maximum at $x = 2$ and the next minimum at $x = 7$ . Find $B$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Find the vertical asymptotes of $y = \tan(2x)$ in the interval $(0, \pi)$ .

Separate answers with commas, e.g. 2, -5

Practice 8

A Ferris wheel rider's height in meters after $t$ minutes is $h(t) = -20\cos\left(\dfrac{\pi t}{5}\right) + 25$ . What is the rider's height at $t = \dfrac{10}{3}$ minutes?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.