Trigonometric graphs: sine, cosine and tangent

Trigonometric graphs turn rotation around a circle into repeating patterns. Their shapes are not arbitrary: for a point $(\cos x,\sin x)$ on the unit circle, cosine records the horizontal coordinate, sine records the vertical coordinate, and tangent records their ratio.

This lesson develops the graphs of $y=\sin x$, $y=\cos x$ and $y=\tan x$, then shows how to read and sketch transformations such as

$$y=a\sin\bigl(b(x-c)\bigr)+d.$$

Unless degrees are shown explicitly, angles are measured in radians.

Prerequisites

You should be able to:

• use exact trigonometric values and quadrant signs;
• convert between degrees and radians;
• identify intercepts, turning points and asymptotes;
• apply graph transformations.

Periodic functions

A function is periodic if its outputs repeat after a fixed positive change in the input. If $T$ is a period, then

$$f(x+T)=f(x)$$

for every $x$ in the domain. The fundamental period is the smallest positive such value.

One complete turn is $2\pi$ radians, so sine and cosine return to the same coordinate after $2\pi$:

$$\sin(x+2\pi)=\sin x, \qquad \cos(x+2\pi)=\cos x.$$

Tangent repeats after only half a turn because both coordinates change sign:

$$\tan(x+\pi) =\frac{-\sin x}{-\cos x} =\tan x.$$

Thus sine and cosine have fundamental period $2\pi$, while tangent has fundamental period $\pi$.

The sine graph

The graph of $y=\sin x$ has domain $\mathbb R$ and range $[-1,1]$. Over one period, its five structural points are

$x$	$0$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$
$\sin x$	$0$	$1$	$0$	$-1$	$0$

Join these points with a smooth curve, then repeat the pattern every $2\pi$. The curve crosses the axis at

$$x=n\pi, \qquad n\in\mathbb Z.$$

It has maximum value $1$ and minimum value $-1$. Sine is an odd function:

$$\sin(-x)=-\sin x,$$

so its graph has rotational symmetry of order two about the origin.

Worked example 1: sketch sine from key points

Sketch $y=\sin x$ for $-\pi\leq x\leq2\pi$.

Start with the points separated by quarter periods:

$$\begin{array}{c|rrrrrrr} x&-\pi&-\pi/2&0&\pi/2&\pi&3\pi/2&2\pi\\ \hline \sin x&0&-1&0&1&0&-1&0 \end{array}$$

Plot these points and join them smoothly. Do not use straight line segments. At each zero, the graph crosses the axis. It rises through $(0,0)$ because small positive angles have positive sine.

The cosine graph

The graph of $y=\cos x$ also has domain $\mathbb R$, range $[-1,1]$ and period $2\pi$.

$x$	$0$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$
$\cos x$	$1$	$0$	$-1$	$0$	$1$

Its roots occur at

$$x=\frac\pi2+n\pi, \qquad n\in\mathbb Z.$$

Cosine is an even function:

$$\cos(-x)=\cos x,$$

so its graph is symmetric about the $y$ axis. Sine and cosine have the same shape, shifted horizontally:

$$\cos x=\sin\left(x+\frac\pi2\right).$$

The plus sign inside shifts the sine graph left by $\pi/2$.

Self check 1

Without a calculator, state:

1. the period and range of $y=\cos x$;
2. the coordinates of the maximum of $y=\sin x$ in $0\leq x\leq2\pi$;
3. all roots of $\cos x$;
4. whether $\sin x$ is even, odd, or neither.

Answers

1. Period $2\pi$ and range $[-1,1]$.
2. $(\pi/2,1)$.
3. $x=\pi/2+n\pi$, where $n\in\mathbb Z$.
4. Odd, because $\sin(-x)=-\sin x$.

The tangent graph

Since

$$\tan x=\frac{\sin x}{\cos x},$$

tangent is undefined wherever $\cos x=0$. These inputs give vertical asymptotes:

$$x=\frac\pi2+n\pi, \qquad n\in\mathbb Z.$$

Between consecutive asymptotes, $y=\tan x$ increases from $-\infty$ to $+\infty$ and crosses the axis at

$$x=n\pi.$$

Its domain and range are

$$x\in\mathbb R\setminus\left\{\frac\pi2+n\pi:n\in\mathbb Z\right\}, \qquad y\in\mathbb R.$$

Tangent has period $\pi$ and is odd because

$$\tan(-x)=-\tan x.$$

An asymptote is not part of the graph. The tangent curve approaches it without touching or crossing it.

Worked example 2: sketch tangent safely

Sketch $y=\tan x$ for $-\pi\leq x\leq\pi$.

First mark the asymptotes

$$x=-\frac\pi2 \qquad\text{and}\qquad x=\frac\pi2.$$

Then mark the roots $(-\pi,0)$, $(0,0)$ and $(\pi,0)$. Useful guide points are

$$\tan\left(\frac\pi4\right)=1, \qquad \tan\left(-\frac\pi4\right)=-1.$$

Draw a separate increasing branch between each pair of asymptotes. Never join branches across an asymptote.

Tangent at an asymptote

$\tan(\pi/2)$ is undefined, not infinity. The symbols $+\infty$ and $-\infty$ describe how nearby values behave; they are not function values.

Transforming sine and cosine

For

$$y=a\sin\bigl(b(x-c)\bigr)+d$$

or the corresponding cosine function, where $a\ne0$ and $b\ne0$:

Feature	Value
amplitude	$
period	$\dfrac{2\pi}{
phase shift	$c$ units right
midline	$y=d$
range	$d-

The amplitude is the maximum vertical distance from the midline, not the full height from minimum to maximum. That full height is $2|a|$.

The factor $b$ acts horizontally, so the period is divided by $|b|$. The shift must be read after factorising the whole argument. For example,

$$\sin(2x-\pi)=\sin\left(2\left(x-\frac\pi2\right)\right),$$

so the shift is $\pi/2$ right, not $\pi$ right.

A reliable sketching method

For one cycle of transformed sine or cosine:

1. Calculate the period $T=2\pi/|b|$.
2. Divide it into four equal steps of length $T/4$.
3. Begin at the phase shift $x=c$.
4. Use the familiar five output levels, transformed by $y\mapsto ay+d$.

For positive $a$, sine uses the pattern

$$d,\quad d+a,\quad d,\quad d-a,\quad d,$$

while cosine uses

$$d+a,\quad d,\quad d-a,\quad d,\quad d+a.$$

The formula still handles $a<0$: multiplying by a negative number reflects the curve in its midline.

Worked example 3: analyse and sketch a sine transformation

For

$$y=3\sin(2x)-1,$$

state the amplitude, period, midline and range, then give five points for one cycle.

Here $a=3$, $b=2$, $c=0$ and $d=-1$. Therefore

$$\text{amplitude}=3, \qquad T=\frac{2\pi}{2}=\pi, \qquad \text{midline }y=-1.$$

The range is

$$-1-3\leq y\leq-1+3,$$

so

$$\boxed{-4\leq y\leq2}.$$

The quarter period is $\pi/4$. Starting at $x=0$, the five points are

$$(0,-1),\quad \left(\frac\pi4,2\right),\quad \left(\frac\pi2,-1\right),\quad \left(\frac{3\pi}4,-4\right),\quad (\pi,-1).$$

These determine one complete cycle.

Worked example 4: extract a hidden phase shift

Describe

$$y=-2\cos(3x+\pi)+4.$$

Factorise the argument:

$$3x+\pi=3\left(x+\frac\pi3\right).$$

Thus the graph is shifted $\pi/3$ left. Its amplitude is $2$, its period is

$$T=\frac{2\pi}{3},$$

and its midline is $y=4$. The negative coefficient reflects the cosine curve in the midline. Its range is

$$4-2\leq y\leq4+2,$$

or

$$\boxed{2\leq y\leq6}.$$

At the phase shift $x=-\pi/3$, the original cosine input is zero. Since $\cos0=1$,

$$y=-2(1)+4=2.$$

The transformed cycle therefore begins at a minimum, not a maximum.

Worked example 5: find a formula from graph features

A sinusoidal graph has maximum $7$, minimum $-1$, period $6$ and crosses its midline at $x=2$ while rising. Find a possible sine formula.

The midline is halfway between the extrema:

$$d=\frac{7+(-1)}2=3.$$

The amplitude is half their difference:

$$a=\frac{7-(-1)}2=4.$$

For period $6$,

$$\frac{2\pi}{|b|}=6 \quad\Longrightarrow\quad |b|=\frac\pi3.$$

A positive sine curve rises through its midline when its argument is zero. The crossing at $x=2$ therefore gives $c=2$. One suitable formula is

$$\boxed{y=4\sin\left(\frac\pi3(x-2)\right)+3}.$$

Equivalent cosine formulae are also possible. A graph does not usually have a unique trigonometric representation.

Transforming tangent

For

$$y=a\tan\bigl(b(x-c)\bigr)+d,$$

the period is

$$\boxed{T=\frac\pi{|b|}}.$$

There is no amplitude because tangent is unbounded. The centre points of its branches lie on the midline $y=d$ at

$$x=c+nT.$$

Vertical asymptotes occur half a period on either side:

$$x=c+\frac{(2n+1)\pi}{2b}, \qquad n\in\mathbb Z.$$

If $a/b>0$, each branch rises from left to right. If $a/b<0$, each branch falls. This sign test accounts for reflections caused by either coefficient.

Worked example 6: transformed tangent

For

$$y=2\tan\left(3\left(x-\frac\pi6\right)\right)-1,$$

state the period, a central point, and the adjacent asymptotes.

The period is

$$T=\frac\pi3.$$

At $x=\pi/6$, the tangent input is zero, so

$$y=2\tan0-1=-1.$$

Thus $(\pi/6,-1)$ is the central point. The adjacent asymptotes are half a period away:

$$x=\frac\pi6\pm\frac\pi6,$$

giving

$$\boxed{x=0\quad\text{and}\quad x=\frac\pi3}.$$

Between them, the branch rises through $(\pi/6,-1)$.

Degrees and radians

The same shapes apply in degrees, but the basic periods change:

Function	Period in radians	Period in degrees
$\sin x$	$2\pi$	$360^\circ$
$\cos x$	$2\pi$	$360^\circ$
$\tan x$	$\pi$	$180^\circ$

Therefore $y=\sin(bx)$ has period $360^\circ/|b|$ when $x$ is measured in degrees, not $2\pi/|b|$.

Worked example 7: a graph in degrees

State the period and range of

$$y=5-2\cos(4x),$$

where $x$ is measured in degrees.

The period is

$$\frac{360^\circ}{4}=90^\circ.$$

The midline is $y=5$ and the amplitude is $2$, so

$$\boxed{3\leq y\leq7}.$$

Common misconceptions

• Using $b$ as the period. For sine and cosine, the period is $2\pi/|b|$; for tangent it is $\pi/|b|$.
• Calling the peak to trough height the amplitude. Amplitude is half that height.
• Reading an inside shift before factorising. Rewrite $bx+k$ as $b(x-c)$ first.
• Giving tangent an amplitude or maximum. Tangent has neither because its range is all real numbers.
• Drawing through a tangent asymptote. Each branch is separate, and the function is undefined on every asymptote.
• Mixing angle units. A radian graph is labelled with multiples of $\pi$; a degree graph is labelled with degree symbols.
• Trusting a calculator window blindly. A narrow or poorly scaled window can hide periods, turning points, or asymptotes. Establish the features algebraically first.

Final self check

1. For $y=-4\sin(3x)+2$, find the amplitude, period, midline and range.
2. For $y=\cos(2x-\pi)$, state the phase shift and give five key $x$ coordinates for one cycle.
3. For $y=\tan(x+\pi/4)$, state the period, central root and adjacent asymptotes.
4. A cosine graph has midline $y=-2$, amplitude $5$, period $\pi$, and a maximum at $x=\pi/3$. Write a possible formula.
5. Explain why $\tan x$ has period $\pi$ although sine and cosine have period $2\pi$.

Answers

1. Amplitude $4$, period $2\pi/3$, midline $y=2$, range $-2\leq y\leq6$.
2. Since $2x-\pi=2(x-\pi/2)$, the shift is $\pi/2$ right. The period is $\pi$, so the quarter period is $\pi/4$. The five coordinates are $x=\pi/2,3\pi/4,\pi,5\pi/4,3\pi/2$.
3. Period $\pi$. The central root is $x=-\pi/4$. The adjacent asymptotes are $x=-3\pi/4$ and $x=\pi/4$.
4. One possible formula is $y=5\cos\bigl(2(x-\pi/3)\bigr)-2$.
5. Increasing $x$ by $\pi$ changes both sine and cosine to their negatives, so their ratio is unchanged: $(-\sin x)/(-\cos x)=\tan x$.

Next steps

Use these graphs to understand principal values in reciprocal and inverse trigonometric functions, then find every solution in an interval in trigonometric equations. The symmetry and periodicity developed here also support trigonometric identities and harmonic form.