What is the Pythagorean identity?

For every angle α, sin²α + cos²α = 1. The notation sin²α means (sin α)², the value of the sine squared — not the sine of the angle squared.

Where does sin²α + cos²α = 1 come from?

It is the Pythagorean theorem applied to a triangle on the unit circle. The point (cos α, sin α) lies on a circle of radius 1, and its coordinates are the legs of a right triangle whose hypotenuse is 1. Hence cos²α + sin²α = 1².

How do you find the cosine from the sine?

Rearrange the identity: cos²α = 1 − sin²α, so cos α = ±√(1 − sin²α). For sin α = 3/5 we get cos²α = 1 − 9/25 = 16/25, so cos α = 4/5 or −4/5. The quadrant the angle lies in decides the sign.

Why does a ± sign appear when finding the cosine?

Because a square root returns a non-negative value, while the cosine is negative in the second and third quadrants. The sine alone does not pin down one angle, which is why a question always states a range such as 0° < α < 90° or 90° < α < 180°.

Is sin²α the same as sin(α²)?

No. sin²α means (sin α)² — take the sine first, then square it. sin(α²) would be the sine of the angle squared, a completely different number. For α = 30°, sin²30° = 0.25 while sin(900°) = 0.

Does the identity hold for obtuse angles?

Yes, for every angle without exception — including negative ones and ones past a full turn. It follows from the equation of the circle x² + y² = 1, which every point of the unit circle satisfies, whatever quadrant it sits in.

What is 1 + tan²α = 1/cos²α used for?

It moves between the tangent and the cosine without computing the sine. It comes from dividing the Pythagorean identity through by cos²α, and it is handy whenever a problem hands you the tangent.

How can I check whether a pair of sine and cosine values is possible?

Square both and add — the result must be exactly 1. The pair sin α = 0.6 and cos α = 0.8 passes (0.36 + 0.64 = 1); the pair 0.5 and 0.5 does not, since it gives 0.5. It is the fastest way to catch an arithmetic slip.

Advanced level

The Pythagorean identity

Q: Is sin²α the same as sin(α²)?

No. sin²α means (sin α)² — take the sine first, then square it. sin(α²) would be the sine of the angle squared, a completely different number. For α = 30°, sin²30° = 0.25 while sin(900°) = 0.

Q: Does the identity hold for obtuse angles?

Yes, for every angle without exception — including negative ones and ones past a full turn. It follows from the equation of the circle x² + y² = 1, which every point of the unit circle satisfies, whatever quadrant it sits in.

Q: What is 1 + tan²α = 1/cos²α used for?

It moves between the tangent and the cosine without computing the sine. It comes from dividing the Pythagorean identity through by cos²α, and it is handy whenever a problem hands you the tangent.

Q: How can I check whether a pair of sine and cosine values is possible?

Square both and add — the result must be exactly 1. The pair sin α = 0.6 and cos α = 0.8 passes (0.36 + 0.64 = 1); the pair 0.5 and 0.5 does not, since it gives 0.5. It is the fastest way to catch an arithmetic slip.

sin²α + cos²α = 1 is the Pythagorean theorem written on the unit circle. See where it comes from, how to find the cosine from the sine, and why an answer is incomplete without knowing the quadrant.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

The unit circleThe unit circle carries the sine and the cosine from a triangle onto the whole circle. See how the coordinates of a point become the values of the functions, what signs they take in the four quadrants, and why the radian is the natural measure of an angle.
The Pythagorean theoremIn a right triangle a² + b² = c². See why it works, how to find the hypotenuse and a leg, what Pythagorean triples are, and how the converse lets you check whether a corner really is square.

All formulas

The Pythagorean identity
$\sin^2 \alpha + \cos^2 \alpha = 1$
holds for every angle, with no exceptions
Sine from cosine
$\sin \alpha = \pm\sqrt{1 - \cos^2 \alpha}$
the quadrant picks the sign
Cosine from sine
$\cos \alpha = \pm\sqrt{1 - \sin^2 \alpha}$
the same identity solved for the cosine
Tangent as a quotient
$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
requires cos α ≠ 0
A derived identity
$1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}$
the identity divided through by cos²α

The sine and the cosine are not independent of each other. Knowing one, you can compute the other — because both are coordinates of the same point on a circle of radius $1$ .

The Pythagorean theorem on a circle

The point $P = (\cos \alpha, \sin \alpha)$ lies on the unit circle. Its $x$ -coordinate, its $y$ -coordinate and the radius form a right triangle:

The legs of this triangle are |cos α| and |sin α| long, and its hypotenuse is exactly 1.

The legs are $\cos \alpha$ and $\sin \alpha$ , and the hypotenuse is the radius $1$ . The Pythagorean theorem therefore gives

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

that is, the Pythagorean identity:

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

This is the equation of the circle $x^2 + y^2 = 1$ written with functions instead of coordinates. That is why it holds for every angle: negative, obtuse, past a full turn. The point always lies on the same circle.

A note on notation: $\sin^2 \alpha$ means $(\sin \alpha)^2$ — sine first, square second. It is an abbreviation, not the sine of the angle squared.

Check the Pythagorean identity for the angle 30°.

Finding the other function

The identity is rearranged like any other equation:

\cos^2 \alpha = 1 - \sin^2 \alpha \quad \Longrightarrow \quad \cos \alpha = \pm\sqrt{1 - \sin^2 \alpha}

\sin^2 \alpha = 1 - \cos^2 \alpha \quad \Longrightarrow \quad \sin \alpha = \pm\sqrt{1 - \cos^2 \alpha}

The $\pm$ is not decoration. A square root returns a non-negative number, while the cosine is sometimes negative — and the sine alone does not identify one angle, since $\sin 30^\circ$ and $\sin 150^\circ$ are the same number although their cosines differ in sign.

Given sin α = 3/5 and 0° < α < 90°, find cos α.

The quadrant picks the sign

The same arithmetic with a different fact about the angle ends in a different answer.

quadrant	range	$\sin \alpha$	$\cos \alpha$
I	$0^\circ < \alpha < 90^\circ$	$+$	$+$
II	$90^\circ < \alpha < 180^\circ$	$+$	$-$
III	$180^\circ < \alpha < 270^\circ$	$-$	$-$
IV	$270^\circ < \alpha < 360^\circ$	$-$	$+$

Given sin α = 3/5 and 90° < α < 180°, find cos α.

The tangent and a derived identity

The tangent is expressed through the other two:

\tan \alpha = \frac{\sin \alpha}{\cos \alpha}

Dividing the Pythagorean identity through by $\cos^2 \alpha$ gives an identity that is useful whenever a problem hands you the tangent:

\frac{\sin^2 \alpha}{\cos^2 \alpha} + \frac{\cos^2 \alpha}{\cos^2 \alpha} = \frac{1}{\cos^2 \alpha} \quad \Longrightarrow \quad \tan^2 \alpha + 1 = \frac{1}{\cos^2 \alpha}

The division is only legal for $\cos \alpha \neq 0$ — and where the cosine vanishes (at $90^\circ$ and $270^\circ$ ) there is no tangent anyway.

Given sin α = 3/5 and α acute, find tan α.

A quick sanity check

The identity is also a checking tool. If your solution produced $\sin \alpha$ and $\cos \alpha$ , square both and add:

$0.6$ and $0.8$ → $0.36 + 0.64 = 1$ ✓
$0.5$ and $0.5$ → $0.25 + 0.25 = 0.5$ ✗ — something is wrong.

Exercises

Each question gives one function and a range for the angle, and asks for the other. Type the answer as a fraction, e.g. 4/5 or −4/5; a decimal is accepted too. The range decides the sign — read it before you type.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

sin α = 3/5, 0° < α < 90° → cos α

Common mistakes

Losing the minus sign in quadrants II and III — the root returns a positive value; the quadrant supplies the sign.
Reading $\sin^2 \alpha$ as $\sin(\alpha^2)$ — the square applies to the value of the function, not to the angle.
$\sqrt{1 - \sin^2 \alpha} = 1 - \sin \alpha$ — the root of a difference is not the difference of the roots.
Subtracting without squaring — the formula takes $\sin^2 \alpha$ , not $\sin \alpha$ .
Ignoring the range given in the question — it is what decides which of the two values to pick.
Dividing by $\cos \alpha$ at $\alpha = 90^\circ$ — the cosine is zero there and the tangent does not exist.

Formula card

Topic: The Pythagorean identity

The Pythagorean identity
$\sin^2 \alpha + \cos^2 \alpha = 1$
holds for every angle, with no exceptions
Sine from cosine
$\sin \alpha = \pm\sqrt{1 - \cos^2 \alpha}$
the quadrant picks the sign
Cosine from sine
$\cos \alpha = \pm\sqrt{1 - \sin^2 \alpha}$
the same identity solved for the cosine
Tangent as a quotient
$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
requires cos α ≠ 0
A derived identity
$1 + \tan^2 \alpha = \frac{1}{\cos^2 \alpha}$
the identity divided through by cos²α

The radius, the cosine and the sine form a right triangle with hypotenuse 1. The Pythagorean theorem for that triangle is exactly the identity.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min