Trigonometric functions: definitions, signs and values

In school geometry, sine and cosine live inside a right triangle, so " $\sin 200°$ " makes no sense — no triangle has a $200°$ angle. Class 11 trigonometry fixes this by moving the definitions onto a unit circle: sine and cosine become the coordinates of a point that sweeps around a circle of radius $1$ , and now any real number, including negative angles and angles past $360°$ , has a well-defined sine and cosine. This article gives the unit-circle definitions of all six trigonometric functions, the fundamental identity $\sin^2x+\cos^2x=1$ , and the domain, range, period, and sign of each function in every quadrant — the toolkit you will use for every trigonometric equation and identity from here on.

The core insight

Sine and cosine are defined as the $y$ - and $x$ -coordinates of a point moving anticlockwise around the unit circle through angle $x$ , so $\sin^2x+\cos^2x=1$ holds for every real $x$ , and the coordinate signs of that point fix the sign of every trigonometric function in each quadrant.

The core idea

Place a unit circle (radius $1$ ) centred at the origin. For a real number $x$ , measured anticlockwise from the positive $x$ -axis, let $P(x)$ be the point reached after rotating through angle $x$ (negative $x$ means clockwise). The two basic functions are defined directly as the coordinates of $P$ :

\cos x = \text{the } x\text{-coordinate of } P(x), \qquad \sin x = \text{the } y\text{-coordinate of } P(x)

where: $x$ is any real number (an angle in radians), and $P(x)$ is the corresponding point on the unit circle. This definition needs no triangle at all, so it extends smoothly to angles beyond $90°$ and to negative angles. The remaining four functions are simply built from these two:

\tan x = \frac{\sin x}{\cos x}\ (\cos x\ne 0), \quad \cot x = \frac{\cos x}{\sin x}\ (\sin x\ne 0), \quad \sec x = \frac{1}{\cos x}\ (\cos x\ne 0), \quad \csc x = \frac{1}{\sin x}\ (\sin x\ne 0)

where: each ratio is undefined exactly where its denominator is zero — $\tan x, \sec x$ fail wherever $\cos x = 0$ , and $\cot x, \csc x$ fail wherever $\sin x = 0$ .

Figure

A unit circle with a point P on it in the second quadrant, its x coordinate labelled cos x and its y coordinate labelled sin x, and the four quadrants marked with which functions are positive in each

Domain, range, period. Since $P(x)$ always lies on the circle, its coordinates never leave $[-1,1]$ , which fixes the range of sine and cosine; the reciprocal and quotient functions inherit their own ranges and are undefined where a denominator vanishes.

Function	Domain	Range	Period
$\sin x$	$\mathbb{R}$	$[-1,1]$	$2\pi$
$\cos x$	$\mathbb{R}$	$[-1,1]$	$2\pi$
$\tan x$	$\mathbb{R} - \{(2n+1)\frac{\pi}{2}\}$	$\mathbb{R}$	$\pi$
$\cot x$	$\mathbb{R} - \{n\pi\}$	$\mathbb{R}$	$\pi$
$\sec x$	$\mathbb{R} - \{(2n+1)\frac{\pi}{2}\}$	$(-\infty,-1]\cup[1,\infty)$	$2\pi$
$\csc x$	$\mathbb{R} - \{n\pi\}$	$(-\infty,-1]\cup[1,\infty)$	$2\pi$

( $n$ ranges over all integers.) $\tan x$ and $\cot x$ repeat twice as often as the other four because rotating $P(x)$ by $\pi$ (half a turn) flips the sign of both $\sin x$ and $\cos x$ together, so their ratio is unchanged.

Signs by quadrant follow directly from the coordinate signs of $P(x)$ : in Quadrant I both coordinates are positive, so all six functions are positive there; in Quadrant II only $y$ (hence $\sin,\csc$ ) is positive; in Quadrant III both coordinates are negative, so only their ratio, $\tan,\cot$ , is positive; in Quadrant IV only $x$ (hence $\cos,\sec$ ) is positive. The mnemonic "All Silver Tea Cups" (ASTC) names which functions are positive in quadrants I, II, III, IV respectively.

Derivation AThe fundamental identity sin squared x plus cos squared x equals 1

The goal is to show, directly from the unit-circle definitions above and nothing else, that $\sin^2x+\cos^2x=1$ holds not just for acute angles in a right triangle but for every real number $x$ , including negative angles and angles beyond $360°$ — this identity is the single most-used tool in the rest of trigonometry, so it must rest on the general definition, not on a triangle that only exists for $0°<x<90°$ .

Step 1. Recall the definition: $P(x) = (\cos x, \sin x)$ is the point reached by rotating anticlockwise through angle $x$ from $(1,0)$ on a circle of radius $1$ centred at the origin.

Step 2. State the equation of that circle. A point $(X,Y)$ lies on a circle of radius $1$ centred at the origin exactly when its distance from the origin is $1$ , i.e. $X^2+Y^2=1^2=1$ . 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 $x$ -axis.

Step 3. Substitute the coordinates of $P(x)$ . Since $X=\cos x$ and $Y=\sin x$ for the specific point $P(x)$ , and $P(x)$ lies on this circle by construction (rotation preserves distance from the centre), the circle's equation becomes

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

Step 4. Note why this holds for every real $x$ . Nothing in Step 3 assumed $x$ was acute, or even positive — $P(x)$ is defined by rotation for any real $x$ , and it always stays on the same unit circle, so the identity is universal, unlike a triangle-based derivation which breaks down outside $0°$ to $90°$ .

Worked example

Problem. Given $\sin x = \dfrac{3}{5}$ and $x$ lies in the second quadrant, find $\cos x$ , $\tan x$ , $\sec x$ , and $\csc x$ .

Step 1. Apply the fundamental identity to get the magnitude of $\cos x$ :

\cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \quad\Rightarrow\quad \cos x = \pm\frac{4}{5}

Step 2. Fix the sign using the quadrant. In Quadrant II, ASTC says only sine and cosecant are positive, so cosine must be negative here:

\cos x = -\frac{4}{5}

Step 3. Compute $\tan x = \dfrac{\sin x}{\cos x}$ and $\sec x = \dfrac{1}{\cos x}$ :

\tan x = \frac{3/5}{-4/5} = -\frac{3}{4}, \qquad \sec x = \frac{1}{-4/5} = -\frac{5}{4}

Step 4. Compute $\csc x = \dfrac{1}{\sin x}$ , which needs no sign decision since $\sin x$ was already given as positive:

\csc x = \frac{1}{3/5} = \frac{5}{3}

Final answer: $\cos x = -\dfrac{4}{5}$ , $\tan x = -\dfrac{3}{4}$ , $\sec x = -\dfrac{5}{4}$ , $\csc x = \dfrac{5}{3}$ .

Competitive stretch

Determine the sign of $\tan\left(\dfrac{5\pi}{4}\right)$ and identify its quadrant, without a calculator.

Step 1. Convert to degrees: $\dfrac{5\pi}{4} \times \dfrac{180}{\pi} = 225°$ . Since $180°<225°<270°$ , the angle lies in Quadrant III.

Step 2. Apply ASTC for Quadrant III: only tan and cot are positive there, so $\tan\left(\dfrac{5\pi}{4}\right) > 0$ . (Its exact value, from the reference angle $\pi/4$ , is $\tan\frac{5\pi}{4}=1$ , consistent with this sign.)

Common traps

Treating $\tan, \cot, \sec, \csc$ as defined everywhere — they are undefined at infinitely many points (wherever $\cos x$ or $\sin x$ equals zero), not just at one or two values.
Mixing up periods: $\sin x$ and $\cos x$ repeat every $2\pi$ , but $\tan x$ and $\cot x$ repeat every $\pi$ — half the period.
Using $\sin^2x+\cos^2x=1$ to find only the magnitude of a ratio and forgetting to fix its sign separately using the quadrant.
Misreading ASTC: it is "All positive in QI, then Sine, Tangent, Cosine positive respectively in QII, QIII, QIV" — confusing which function goes with which quadrant is a common slip.

Ask the tutor: Want a quick drill on identifying the sign and quadrant of angles like $\dfrac{7\pi}{6}$ or $-40°$ ?

Why does the unit-circle definition work for angles bigger than 360 degrees or negative angles, when the old right-triangle definition does not?

A right triangle only has angles strictly between $0°$ and $90°$ , so " $\sin 400°$ " or " $\sin(-30°)$ " simply cannot be built from a triangle. The unit-circle definition instead uses rotation: an angle of $400°$ just means you keep rotating anticlockwise past a full turn ( $360°$ ) and go another $40°$ , landing on the same point as $40°$ . A negative angle means rotating clockwise instead. Since rotation by any real number of degrees or radians always lands $P(x)$ somewhere on the circle, $\cos x$ and $\sin x$ are defined for every real $x$ , with no upper or lower limit.

Do I need to memorise the domain and range of all six functions separately, or can I derive them?

You can derive almost all of it from just two facts: $\sin x$ and $\cos x$ never leave $[-1,1]$ (coordinates on a unit circle), and a ratio is undefined exactly where its denominator is zero. From this, $\tan x=\sin x/\cos x$ and $\sec x=1/\cos x$ fail wherever $\cos x=0$ , i.e. at $x=(2n+1)\pi/2$ ; similarly $\cot x$ and $\csc x$ fail wherever $\sin x=0$ , i.e. at $x=n\pi$ . The ranges of $\sec x,\csc x$ being $(-\infty,-1]\cup[1,\infty)$ follow because they are reciprocals of something in $[-1,1]$ excluding zero.

If sin x is given, can cos x always be found using the identity alone, without knowing the quadrant?

Only up to a sign. The identity $\sin^2x+\cos^2x=1$ gives $\cos x = \pm\sqrt{1-\sin^2x}$ — it fixes the size of $\cos x$ but leaves two possibilities. To choose between them you need one extra piece of information, almost always which quadrant $x$ lies in, since that's what the ASTC rule uses to decide the sign. Without knowing the quadrant (or an equivalent clue, like the sign of another ratio), the answer for $\cos x$ genuinely stays ambiguous between two values.

Trigonometric functions: definitions, signs and values

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads