Trigonometric functions

Sine and cosine start life as ratios in a right triangle, useful only for acute angles below $90^\circ$ . Trigonometric functions break that ceiling. By reading the coordinates of a point moving around a circle of radius $1$ , sine and cosine become functions of any real number — negative, zero, or larger than a full turn — and the other four ratios follow from them. This single idea, the unit-circle definition, is the spine of the whole chapter: it delivers the fundamental identity $\sin^2x+\cos^2x=1$ for free, fixes the exact domain and range of all six functions, and explains why some of them repeat every $2\pi$ while others repeat every $\pi$ . Everything that follows — signs in each quadrant, evaluating $\sin(7\pi/6)$ , solving equations — is this one picture used again and again.

The core insight

Defining $\cos x$ and $\sin x$ as the coordinates of a point reached by travelling an arc length $x$ around the unit circle extends trigonometry from acute-angle ratios to functions of every real number, with $\sin^2x+\cos^2x=1$ following directly from the circle's equation.

The core idea

From a triangle ratio to a circle picture. In a right triangle, $\sin\theta$ and $\cos\theta$ only make sense for $0^\circ<\theta<90^\circ$ . To go beyond that, take a circle of radius $1$ centred at the origin $O$ — the unit circle — and a starting point $P_0=(1,0)$ on it. For a real number $x$ , walk along the circle through an arc length $x$ , anticlockwise if $x>0$ and clockwise if $x<0$ , to land on a point $P(a,b)$ . Define

\cos x = a, \qquad \sin x = b.

where: $x$ is the arc length travelled, measured in radians (so $x$ doubles as the angle $\angle P_0OP$ ), and $(a,b)$ are the coordinates of the point reached. Because the radius is $1$ , arc length and radian angle are the same number, so this matches the old right-triangle sine and cosine for acute $x$ while also making sense for $x=0$ , negative $x$ , or $x$ bigger than $2\pi$ (multiple loops around the circle). The remaining four functions are ratios built from sine and cosine, defined wherever the denominator does not vanish:

\tan x = \frac{\sin x}{\cos x}\ (\cos x\neq0), \qquad \cot x = \frac{\cos x}{\sin x}\ (\sin x\neq0),

\sec x = \frac{1}{\cos x}\ (\cos x\neq0), \qquad \operatorname{cosec} x = \frac{1}{\sin x}\ (\sin x\neq0).

where: each function is undefined exactly where its denominator is zero — a fact worth memorising because it fixes the domain of four of the six functions.

Figure

A unit circle with a point P at arc length x showing cosine as the horizontal coordinate and sine as the vertical coordinate

Domain, range, and period. Because $a^2+b^2=1$ on the unit circle, both $\sin x$ and $\cos x$ take every value in $[-1,1]$ and are defined for all real $x$ — there is no gap or forbidden point to worry about for these two. $\tan x$ and $\sec x$ break down wherever $\cos x=0$ , that is at $x=(2n+1)\tfrac{\pi}{2}$ ; $\cot x$ and $\operatorname{cosec} x$ break down wherever $\sin x=0$ , that is at $x=n\pi$ ( $n\in\mathbb{Z}$ ). Their ranges are $\mathbb{R}$ for tangent and cotangent, and $\mathbb{R}-(-1,1)$ for secant and cosecant, since dividing $1$ by a number in $[-1,1]$ (excluding $0$ ) always gives something of size at least $1$ : as $\cos x$ shrinks toward $0$ , $\sec x=1/\cos x$ grows without bound, so no value strictly between $-1$ and $1$ can ever occur. Sine and cosine (and therefore secant and cosecant) repeat every $2\pi$ , one full lap of the circle; tangent and cotangent, because a half-turn sends both coordinates to their negatives and their ratio is unchanged, already repeat after just $\pi$ , half a lap — the shortest interval after which every one of the six functions can be guaranteed to return to its starting value is therefore $2\pi$ , but tangent and cotangent settle back sooner.

Worked example

Problem. If $\cos x=-\dfrac{3}{5}$ and $x$ lies in the third quadrant, find $\sin x$ , $\tan x$ , and, as a stretch, $\sec x$ , $\operatorname{cosec} x$ , and $\cot x$ .

Step 1. Use the fundamental identity to get $\sin^2x$ : $\sin^2x=1-\cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}$ , so $\sin x=\pm\dfrac{4}{5}$ .

Step 2. Fix the sign from the quadrant. In the third quadrant both sine and cosine are negative (only tangent and cotangent are positive there), matching the given negative $\cos x=-\dfrac{3}{5}$ . So take the negative root: $\sin x=-\dfrac{4}{5}$ .

Step 3. Compute $\tan x=\dfrac{\sin x}{\cos x}=\dfrac{-4/5}{-3/5}=\dfrac{4}{3}$ .

Competitive stretch

Step 4. The remaining three functions are reciprocals of what we already have: $\sec x=\dfrac{1}{\cos x}=-\dfrac{5}{3}$ , $\operatorname{cosec} x=\dfrac{1}{\sin x}=-\dfrac{5}{4}$ , and $\cot x=\dfrac{1}{\tan x}=\dfrac{3}{4}$ . As a quick check, $1+\tan^2x=1+\dfrac{16}{9}=\dfrac{25}{9}=\sec^2x$ , which matches $\left(-\dfrac{5}{3}\right)^2=\dfrac{25}{9}$ .

Final answer: $\sin x=-\dfrac{4}{5}$ , $\tan x=\dfrac{4}{3}$ , $\sec x=-\dfrac{5}{3}$ , $\operatorname{cosec} x=-\dfrac{5}{4}$ , $\cot x=\dfrac{3}{4}$ .

Derivation AThe fundamental identity from the unit circle

The goal is to prove, using nothing beyond the equation of a circle and the unit-circle definition of sine and cosine, that $\sin^2x+\cos^2x=1$ holds for every real number $x$ , and then to turn that one identity into the two companion identities $1+\tan^2x=\sec^2x$ and $1+\cot^2x=\operatorname{cosec}^2x$ by simple division, so a student can see all three come from a single geometric fact rather than three separate rules to memorise.

Step 1. Write down the equation of the unit circle. Every point $(a,b)$ on a circle of radius $1$ centred at the origin satisfies $a^2+b^2=1$ — this is just the Pythagorean theorem applied to the right triangle formed by the point, the origin, and the foot of the perpendicular to the horizontal axis, with hypotenuse equal to the radius $1$ .

Step 2. Substitute the unit-circle definition. By definition, the point $P$ reached after arc length $x$ has coordinates $(a,b)=(\cos x,\sin x)$ . Substituting these into $a^2+b^2=1$ gives immediately

\sin^2x+\cos^2x=1

for every real $x$ , since the construction of $P$ places it on the unit circle no matter how large or small $x$ is, or which direction it is measured.

Step 3. Divide by $\cos^2x$ to get the tangent–secant identity. Wherever $\cos x\neq0$ , dividing every term of $\sin^2x+\cos^2x=1$ by $\cos^2x$ gives $\dfrac{\sin^2x}{\cos^2x}+1=\dfrac{1}{\cos^2x}$ , which is exactly $\tan^2x+1=\sec^2x$ , using the ratio definitions of $\tan x$ and $\sec x$ from above.

Step 4. Divide by $\sin^2x$ to get the cotangent–cosecant identity. Wherever $\sin x\neq0$ , dividing every term of $\sin^2x+\cos^2x=1$ by $\sin^2x$ gives $1+\dfrac{\cos^2x}{\sin^2x}=\dfrac{1}{\sin^2x}$ , that is $1+\cot^2x=\operatorname{cosec}^2x$ . All three identities are thus one statement — the Pythagorean theorem on the unit circle — viewed three ways.

Common traps

Guessing the sign instead of using the quadrant. Taking a square root from $\sin^2x+\cos^2x=1$ always gives two possible signs; the quadrant stated in the problem, not a default choice of positive, decides which one is correct.
Ignoring domain restrictions. $\tan x$ and $\sec x$ are undefined at odd multiples of $\dfrac{\pi}{2}$ ; $\cot x$ and $\operatorname{cosec} x$ are undefined at multiples of $\pi$ . Plugging these values in anyway is a common slip.
Mixing up the two periods. Sine and cosine repeat every $2\pi$ , but tangent and cotangent repeat every $\pi$ — using $2\pi$ where $\pi$ is correct throws off simplification and equation-solving questions.
Treating the range of secant and cosecant as excluding just zero. Their range excludes the entire open interval $(-1,1)$ , not only the single point $0$ .

Ask the tutor: want to see the unit-circle picture animated, or practice evaluating a few more angles like $\dfrac{7\pi}{6}$ or $\dfrac{5\pi}{3}$ step by step?

Why does the unit circle need radius exactly 1?

Radius $1$ is what makes arc length and radian angle the same number, so a single quantity $x$ can serve as both "how far along the circle" and "the angle turned through." With any other radius $r$ , the arc length for angle $x$ radians would be $rx$ , and the definitions $\cos x=a$ , $\sin x=b$ would need an extra division by $r$ to stay consistent. Choosing radius $1$ removes that clutter and keeps the definitions clean.

Why do tan x and cot x repeat every pi instead of every 2 pi like sine and cosine?

Moving halfway around the circle, from $x$ to $x+\pi$ , sends the point $P(a,b)$ to its exact opposite $(-a,-b)$ — both coordinates flip sign. So $\sin(x+\pi)=-\sin x$ and $\cos(x+\pi)=-\cos x$ , meaning sine and cosine individually need a full $2\pi$ to return to their original values. But $\tan(x+\pi)=\dfrac{-\sin x}{-\cos x}=\dfrac{\sin x}{\cos x}=\tan x$ : the two sign flips cancel in the ratio, so tangent (and by the same argument cotangent) already returns to its original value after just $\pi$ .

How do I quickly find sin, cos and tan of angles like 7 pi over 6 without a calculator?

Split the angle into a multiple of $\pi$ (or $\pi/2$ ) plus a small standard angle, then use the sign rule for that quadrant together with the value of the standard angle. For $\dfrac{7\pi}{6}=\pi+\dfrac{\pi}{6}$ , the angle $\pi+\theta$ lands in the third quadrant, where sine and cosine are both negative, and the standard results $\sin(\pi+\theta)=-\sin\theta$ , $\cos(\pi+\theta)=-\cos\theta$ let you read off $\sin\dfrac{7\pi}{6}=-\dfrac12$ and $\cos\dfrac{7\pi}{6}=-\dfrac{\sqrt3}{2}$ straight from the known values at $\theta=\pi/6$ .

Trigonometric functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads