Trigonometric graphs

Six functions, six unmistakable shapes — and once you can draw them from memory, half of trigonometry becomes a picture instead of a formula. $y=\sin x$ and $y=\cos x$ are the same smooth wave, one shifted from the other, forever trapped between $-1$ and $1$ . $y=\tan x$ escapes that cage entirely, shooting off to infinity at regular intervals. Their reciprocals — $\csc x$ , $\sec x$ , $\cot x$ — turn those very same infinities into vertical asymptotes and turn the bounded humps of sine and cosine into U-shaped branches that never come near the $x$ -axis. This article sketches all six graphs over one full period, pins down exactly where each one repeats (its period) and how far it stretches vertically (its amplitude), and shows how the general form $y=a\sin(bx+c)+d$ compresses, stretches, and shifts the basic wave — all without leaving the ISC Class 11 syllabus.

The core insight

Sine and cosine are one smooth, bounded wave with period $2\pi$ and range $[-1,1]$ that repeats every $\frac{2\pi}{b}$ once scaled to $\sin(bx)$ ; tangent and cotangent are unbounded with period $\pi$ ; and cosecant, secant repeat every $2\pi$ but never enter the strip $-1<y<1$ , each blowing up exactly where its reciprocal partner is zero.

The core idea

$y=\sin x$ . A smooth wave through the origin: $\sin 0=0$ , rising to a maximum $\sin\frac{\pi}{2}=1$ , back to $0$ at $\pi$ , down to a minimum $\sin\frac{3\pi}{2}=-1$ , and returning to $0$ at $2\pi$ . Period $2\pi$ , range $[-1,1]$ , amplitude $1$ .

$y=\cos x$ . Identical wave shape to sine but shifted — it starts at its maximum $\cos 0=1$ , crosses $0$ at $\frac{\pi}{2}$ , hits minimum $-1$ at $\pi$ , back to $0$ at $\frac{3\pi}{2}$ , and returns to $1$ at $2\pi$ . In fact $\cos x=\sin\left(x+\frac{\pi}{2}\right)$ : cosine is sine shifted left by $\frac{\pi}{2}$ . Period $2\pi$ , range $[-1,1]$ .

$y=\tan x$ . Unbounded: it rises from $-\infty$ to $+\infty$ on each interval between consecutive vertical asymptotes at odd multiples of $\frac{\pi}{2}$ (where $\cos x=0$ ), passing through the origin and every branch strictly increasing. Its period is half that of sine and cosine.

\tan(x+\pi)=\tan x

where: $x$ is any real number away from an asymptote, so the graph repeats itself every $\pi$ radians rather than every $2\pi$ — the single most common thing students forget.

Reciprocal trio. $y=\csc x=\frac{1}{\sin x}$ is undefined wherever $\sin x=0$ (integer multiples of $\pi$ ), each sine hump turning into a U- or ∩-shaped branch bulging away from the axis; range $(-\infty,-1]\cup[1,\infty)$ , period $2\pi$ . $y=\sec x=\frac{1}{\cos x}$ behaves the same way but is undefined at odd multiples of $\frac{\pi}{2}$ (where $\cos x=0$ ), so its branches sit where cosecant's do not. $y=\cot x=\frac{1}{\tan x}$ is undefined at multiples of $\pi$ and, unlike $\tan x$ , is decreasing on every branch; period $\pi$ .

Figure

Six trigonometric graphs stacked in a grid showing the smooth bounded waves of sine and cosine, the repeating increasing branches of tangent, and the asymptotic U shaped branches of cosecant, secant and cotangent

Derivation AThe period of y = sin(bx) and y = tan(bx)

The goal is to show, using nothing more than the definition of "period" and the known repeat-length of the parent function, exactly why scaling the input of a trigonometric function by a constant $b$ divides its period by $|b|$ — the rule that explains why $\tan x$ repeats every $\pi$ while $\sin x$ repeats every $2\pi$ , and why compressed graphs like $y=\sin 2x$ finish a full wave twice as fast.

Step 1. Start from the definition: $T$ is a period of $y=f(x)$ if $f(x+T)=f(x)$ for every $x$ in the domain, and the period is the smallest such positive $T$ . For the parent sine function, $\sin(x+2\pi)=\sin x$ for all $x$ , and no smaller positive number works, so $\sin x$ has period $2\pi$ .

Step 2. Consider $y=\sin(bx)$ with $b>0$ . Substituting $x\to x+T$ requires $\sin\big(b(x+T)\big)=\sin(bx)$ for every $x$ , i.e. $\sin(bx+bT)=\sin(bx)$ .

Step 3. Since sine's own period is $2\pi$ , this holds for every $x$ exactly when the extra angle $bT$ equals a full cycle of the parent function, $bT=2\pi$ . Solving, $T=\dfrac{2\pi}{b}$ .

Step 4. Apply the identical argument to $y=\tan(bx)$ , but the parent function $\tan x$ has period $\pi$ , not $2\pi$ , so the matching condition is $bT=\pi$ , giving $T=\dfrac{\pi}{b}$ . Setting $b=1$ recovers the familiar period- $2\pi$ sine/cosine graphs and period- $\pi$ tangent/cotangent graphs described above; setting $b=2$ , for instance, halves each of those periods.

Worked example

Problem. Sketch $y=3\sin(2x)$ for $0\le x\le\pi$ , stating the amplitude, the period, and the coordinates of the maximum and minimum points, then use the graph to state the range.

Step 1. Read off the amplitude and period directly from the form $y=a\sin(bx)$ with $a=3$ , $b=2$ : amplitude $=|a|=3$ , and by the period formula derived above, $T=\dfrac{2\pi}{b}=\dfrac{2\pi}{2}=\pi$ . So the whole graph in this problem covers exactly one full period.

Step 2. Build a table of key values by scaling the standard $y=\sin x$ table from the core idea (using $\sin\frac{\pi}{2}=1$ , $\sin\pi=0$ , $\sin\frac{3\pi}{2}=-1$ ): at $x=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi$ , the angle $2x$ takes the standard values $0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$ , giving $y=3\sin(2x)=0,3,0,-3,0$ .

Step 3. Plot these five points — $(0,0)$ , $\left(\frac{\pi}{4},3\right)$ , $\left(\frac{\pi}{2},0\right)$ , $\left(\frac{3\pi}{4},-3\right)$ , $(\pi,0)$ — and join with one smooth wave, compressed to half the width of $y=\sin x$ and stretched to three times its height.

Step 4. The maximum point is $\left(\frac{\pi}{4},3\right)$ and the minimum point is $\left(\frac{3\pi}{4},-3\right)$ ; since the curve sweeps through every value between these extremes, the range is

[-3,3]

The graph oscillates between $y=-3$ and $y=3$ , completing one full wave over $0\le x\le\pi$ .

Competitive stretch

If instead $y=3\sin(2x)+1$ , every output is shifted up by $1$ : the same wave, same amplitude $3$ , same period $\pi$ , but now oscillating between $-3+1=-2$ and $3+1=4$ , so the range becomes $[-2,4]$ — the " $+d$ " in $y=a\sin(bx+c)+d$ always slides the whole graph vertically without touching its shape.

Common traps

Drawing $\sin x$ or $\cos x$ with sharp corners like a triangle wave instead of a smooth, continuously curving wave.
Forgetting that $\tan x$ and $\cot x$ repeat every $\pi$ , not $2\pi$ — sketching only one hump over $[0,2\pi]$ when two full periods actually occur.
Drawing $\sec x$ or $\csc x$ branches touching or crossing the $x$ -axis, when in fact $|\sec x|\ge1$ and $|\csc x|\ge1$ always, so their graphs never enter the strip $-1<y<1$ .
Mixing up which coefficient does what in $y=a\sin(bx+c)+d$ : only $a$ changes the height (amplitude), only $b$ changes how fast it repeats (period $=2\pi/b$ ), and $d$ merely slides the whole graph up or down.

Ask the tutor: if you sketch $y=2\cos(3x)$ and describe its shape, the AI tutor can check whether your amplitude, period, and turning points are all correctly placed.

Why does tan x have period pi while sin x and cos x have period 2pi?

Because $\tan x=\frac{\sin x}{\cos x}$ , and flipping the sign of both sine and cosine together (which happens after adding $\pi$ , not a full $2\pi$ ) leaves their ratio unchanged. Concretely, $\sin(x+\pi)=-\sin x$ and $\cos(x+\pi)=-\cos x$ , so $\tan(x+\pi)=\frac{-\sin x}{-\cos x}=\tan x$ . The two minus signs cancel, so tangent repeats twice as often as the sine and cosine it is built from. Cotangent, being $1/\tan x$ , inherits the same period $\pi$ .

How do I remember where the asymptotes of sec x and csc x go without memorising two separate lists?

Use the reciprocal relationship directly instead of memorising: $\sec x=\frac{1}{\cos x}$ blows up exactly where $\cos x=0$ , and $\csc x=\frac{1}{\sin x}$ blows up exactly where $\sin x=0$ . So first sketch (or recall) the zeros of cosine and sine from the parent graphs in the core idea — odd multiples of $\frac{\pi}{2}$ for cosine, integer multiples of $\pi$ for sine — and those become your asymptote list automatically.

In y = a sin(bx + c) + d, what does c actually do, and why was it not needed in the worked example?

The constant $c$ produces a horizontal shift (a "phase shift") of the graph by $-\frac{c}{b}$ along the $x$ -axis, sliding the wave left or right without changing its height or width. It was left out of the worked example, and out of most ISC Class 11 sketching questions, because the syllabus at this stage focuses on amplitude ( $a$ ), period ( $b$ ), and vertical shift ( $d$ ); phase-shifted sketches are typically only asked once you have also studied the compound-angle identities that make $c$ easy to interpret.

Trigonometric graphs

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads