The Pythagorean Identity: sin²θ + cos²θ = 1

Quick answer

On the unit circle a point is (cos θ, sin θ) and the circle's equation is x² + y² = 1. Substituting gives cos²θ + sin²θ = 1 in one line. It holds for every angle, which makes it an identity rather than an equation to solve. Its everyday use is filling in a missing value: one of sine or cosine plus the quadrant is enough to pin down the other exactly.

What you'll learn

Prove the Pythagorean identity from the unit circle
Find sine, cosine or tangent given one value and the quadrant
Rewrite expressions using the identity and its two other forms

The circle’s equation, renamed

A circle of radius $1$ centered at the origin has equation

x^2 + y^2 = 1

On the unit circle, the point for an angle $\theta$ is $(\cos\theta,\ \sin\theta)$ . Those are the $x$ and the $y$ . Substitute:

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

That is the whole proof. The identity is the circle’s equation with the coordinates called by their trigonometric names.

The Pythagorean theorem on the unit circle

The picture says the same thing. The dashed vertical makes a right triangle with horizontal leg $\cos\theta$ , vertical leg $\sin\theta$ , and hypotenuse equal to the radius, which is $1$ . The Pythagorean theorem then reads

(\cos\theta)^2 + (\sin\theta)^2 = 1^2

Reading sin²θ

\sin^2\theta \ \text{ means } \ (\sin\theta)^2

The exponent is parked on the function name to keep parentheses out of the way. It never means $\sin(\theta^2)$ , and it is unrelated to $\sin^{-1}$ , which names the inverse function rather than a reciprocal.

Why it holds for every angle

An equation such as $2x + 1 = 7$ is a question: which $x$ makes this true? An identity is a statement: this is true for every value, so there is nothing to solve.

The Pythagorean identity is the second kind, and the unit circle shows why. Any angle at all — acute, obtuse, negative, past a full turn — has a terminal side that meets the circle somewhere. Wherever it meets, the point is on a circle of radius $1$ , so its coordinates satisfy $x^2 + y^2 = 1$ . No angle escapes.

The squares are what make the signs stop mattering. In Quadrant II the first coordinate is negative, but squaring erases the sign, so the sum comes out the same as in Quadrant I.

Squaring erases the quadrant

All four points give $0.36 + 0.64 = 1$ . The identity holds in every quadrant because the quadrant only controls signs, and the squares discard them.

Finding the value you are missing

This is what the identity is used for. Know one of $\sin\theta$ or $\cos\theta$ , know the quadrant, and the other value is pinned down exactly.

The steps are always the same:

Substitute the known value into $\sin^2\theta + \cos^2\theta = 1$ .
Solve for the square of the unknown.
Take the square root, which gives two candidates.
Use the quadrant to choose the sign.

Step 4 is the one that gets skipped, and it is the only step that carries information the algebra cannot supply.

Tangent comes along for free

Once both coordinates are known, tangent is a division.

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

With the values from Example 1:

\tan\theta = \frac{\tfrac{3}{5}}{-\tfrac{4}{5}} = -\tfrac{3}{4}

The sign is worth a second look. Quadrant II has a positive height over a negative width, so tangent is negative there. It matches.

The same identity, divided

Dividing by a square produces two relatives. Start from the identity and divide every term by $\cos^2\theta$ :

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

Dividing by $\sin^2\theta$ instead gives

1 + \cot^2\theta = \csc^2\theta

where the three reciprocal functions are

\sec\theta = \frac{1}{\cos\theta} \qquad \csc\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{1}{\tan\theta}

These are not three facts to memorize. They are one fact and two divisions, and rebuilding either version takes a single line.

Worked examples

Example 6 — Starting from tangent

Given $\tan\theta = -\tfrac{7}{24}$ with $\theta$ in Quadrant II, find $\sin\theta$ and $\cos\theta$ .

Use the divided form.

\sec^2\theta = 1 + \tan^2\theta = 1 + \tfrac{49}{576} = \tfrac{625}{576}

\sec\theta = \pm\tfrac{25}{24}

In Quadrant II cosine is negative, so its reciprocal is too: $\sec\theta = -\tfrac{25}{24}$ , giving

\cos\theta = -\tfrac{24}{25}

Then $\sin\theta = \tan\theta \cdot \cos\theta$ :

\sin\theta = \left(-\tfrac{7}{24}\right)\left(-\tfrac{24}{25}\right) = \tfrac{7}{25}

Check: $\tfrac{49}{625} + \tfrac{576}{625} = \tfrac{625}{625} = 1$ ✓

Common mistakes

Practice problems

Given $\sin\theta = \tfrac{3}{5}$ with $\theta$ in Quadrant I, find $\cos\theta$ .

Answer

$\tfrac{4}{5}$

Full solution

$\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$ , so $\cos\theta = \pm\tfrac{4}{5}$ .

Quadrant I has both coordinates positive, so $\cos\theta = \tfrac{4}{5}$ .
Given $\cos\theta = \tfrac{5}{13}$ with $\theta$ in Quadrant IV, find $\sin\theta$ .

Answer

$-\tfrac{12}{13}$

Full solution

$\sin^2\theta = 1 - \tfrac{25}{169} = \tfrac{144}{169}$ , so $\sin\theta = \pm\tfrac{12}{13}$ .

Quadrant IV is below the $x$ -axis, so the height is negative.
Verify the identity at $\theta = \tfrac{\pi}{4}$ .

Answer

$\tfrac{1}{2} + \tfrac{1}{2} = 1$ ✓

Full solution

Both values are $\tfrac{\sqrt{2}}{2}$ , and $\left(\tfrac{\sqrt{2}}{2}\right)^2 = \tfrac{2}{4} = \tfrac{1}{2}$ .

Adding the two halves gives $1$ .
Given $\sin\theta = -\tfrac{8}{17}$ with $\theta$ in Quadrant III, find $\cos\theta$ .

Answer

$-\tfrac{15}{17}$

Full solution

$\cos^2\theta = 1 - \tfrac{64}{289} = \tfrac{225}{289}$ , so $\cos\theta = \pm\tfrac{15}{17}$ .

Quadrant III has both coordinates negative.
Simplify $1 - \sin^2\theta$ .

Answer

$\cos^2\theta$

Full solution

Rearranging $\sin^2\theta + \cos^2\theta = 1$ gives $\cos^2\theta = 1 - \sin^2\theta$ .
Find $\tan\theta$ for the angle in problem 4.

Answer

$\tfrac{8}{15}$

Full solution

$\tan\theta = \left(-\tfrac{8}{17}\right) \div \left(-\tfrac{15}{17}\right) = \tfrac{8}{15}$

Two negatives divide to a positive, which matches Quadrant III, where tangent is positive.
Given $\cos\theta = -\tfrac{20}{29}$ with $\theta$ in Quadrant II, find $\sin\theta$ and $\tan\theta$ .

Answer

$\sin\theta = \tfrac{21}{29}$ and $\tan\theta = -\tfrac{21}{20}$

Full solution

$\sin^2\theta = 1 - \tfrac{400}{841} = \tfrac{441}{841}$ , so $\sin\theta = \pm\tfrac{21}{29}$ .

Quadrant II is above the $x$ -axis, so $\sin\theta = \tfrac{21}{29}$ .

$\tan\theta = \tfrac{21}{29} \div \left(-\tfrac{20}{29}\right) = -\tfrac{21}{20}$
Given $\sin\theta = \tfrac{1}{4}$ with $\theta$ in Quadrant II, find $\cos\theta$ .

Hint

The answer is not a whole-number fraction. Leave the radical in place.

Answer

$-\tfrac{\sqrt{15}}{4}$

Full solution

$\cos^2\theta = 1 - \tfrac{1}{16} = \tfrac{15}{16}$

$\cos\theta = \pm\tfrac{\sqrt{15}}{4}$

$15$ has no square factor, so the radical stays.

Quadrant II makes the first coordinate negative: $\cos\theta = -\tfrac{\sqrt{15}}{4}$ .
Simplify $(\sin^2\theta + \cos^2\theta) \div \cos\theta$ .

Answer

$\sec\theta$ , which is $\tfrac{1}{\cos\theta}$

Full solution

The numerator is $1$ by the identity, so the expression is $\tfrac{1}{\cos\theta}$ .

That reciprocal has a name: $\sec\theta$ .
Given $\sin\theta = \tfrac{3}{5}$ with $\theta$ in Quadrant II, Dana answers $\cos\theta = \tfrac{4}{5}$ , saying cosine measures a length so it cannot be negative. Find the error.

Hint

Where is the point for a Quadrant II angle, relative to the $y$ -axis?

Answer

Cosine is a coordinate, not a length. $\cos\theta = -\tfrac{4}{5}$ .

Full solution

Dana’s algebra is right as far as it goes: $\cos^2\theta = \tfrac{16}{25}$ , so $\cos\theta = \pm\tfrac{4}{5}$ .

The error is in choosing the sign. Cosine is defined as the first coordinate of a point on the unit circle, and coordinates carry signs. The length in the picture is the radius, which is $1$ .

A Quadrant II angle lands left of the $y$ -axis, where first coordinates are negative.

So $\cos\theta = -\tfrac{4}{5}$ .

A check confirms the point is on the circle: $\left(-\tfrac{4}{5}\right)^2 + \left(\tfrac{3}{5}\right)^2 = \tfrac{16}{25} + \tfrac{9}{25} = 1$ ✓

Frequently asked questions

What is the Pythagorean identity?

sin²θ + cos²θ = 1, true for every angle θ. It is the equation of the unit circle written with trigonometric names.

What does sin²θ mean?

The square of sin θ. The exponent is written on the function name to avoid the clutter of (sin θ)², and it never means the sine of θ².

How do I find cosine when I know sine?

Substitute into the identity and solve for cos²θ, then take the square root. The quadrant decides whether the answer is the positive or the negative root.

Why is a quadrant needed?

The square root gives two candidates that differ only in sign. Knowing the quadrant tells you the sign of the first coordinate, which picks one.

Where do tan²θ + 1 = sec²θ and 1 + cot²θ = csc²θ come from?

Both come from dividing sin²θ + cos²θ = 1 by cos²θ and by sin²θ. They are the same identity in different clothes.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSF.TF.C.8Trigonometric FunctionsProve the Pythagorean identity sin²(θ) + cos²(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Frequently asked questions

What is the Pythagorean identity?

What does sin²θ mean?

How do I find cosine when I know sine?

Why is a quadrant needed?

Where do tan²θ + 1 = sec²θ and 1 + cot²θ = csc²θ come from?

What to learn next

Standards alignment

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