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Pythagorean Identities

Pythagorean Identities: Only Memorize One — Here’s How to Derive the Other Two

Quick Answer — The Three Pythagorean Identities

sin²θ + cos²θ = 1

The foundation — comes directly from the unit circle

tan²θ + 1 = sec²θ

Derived by dividing the first identity by cos²θ

1 + cot²θ = csc²θ

Derived by dividing the first identity by sin²θ

You only need to memorize the first one. The other two are derived from it in two steps each.

Where sin²θ + cos²θ = 1 Comes From

The Pythagorean identity is not a formula someone invented and decided to memorize. It’s a direct consequence of the Pythagorean theorem applied to the unit circle — and understanding that derivation means you’ll never forget it.

Place any point P on a circle of radius 1. By the unit circle definition, P has coordinates (cos θ, sin θ). Draw a right triangle from the center to P: the horizontal leg has length cos θ, the vertical leg has length sin θ, and the hypotenuse has length 1 (the radius). The Pythagorean theorem says the sum of the squares of the two legs equals the square of the hypotenuse. That gives:

Unit circle with a right triangle inside showing horizontal leg cos theta, vertical leg sin theta, and hypotenuse equal to 1. Right panel shows the derivation from x squared plus y squared equals r squared to sin squared theta plus cos squared theta equals 1.

The Pythagorean theorem on a unit circle gives sin²θ + cos²θ = 1 directly. This is geometry, not algebra.

This is why the identity is true for every angle without exception — not just for 30°, 45°, and 60°, but for any angle including non-standard ones. The point P is always on the unit circle, the triangle always has hypotenuse 1, and the Pythagorean theorem always applies.

The Three Pythagorean Identities

Derivation map showing sin squared theta plus cos squared theta equals 1 in the center. Dividing by cos squared theta on the left gives tan squared theta plus 1 equals sec squared theta. Dividing by sin squared theta on the right gives 1 plus cot squared theta equals csc squared theta.

All three identities are the same geometric fact — just divided through by different functions. Memorize only the first; derive the rest.

sin²θ + cos²θ = 1

The primary form

Also: cos²θ = 1 − sin²θ
sin²θ = 1 − cos²θ

tan²θ + 1 = sec²θ

Divide first identity by cos²θ

Also: sec²θ − tan²θ = 1
tan²θ = sec²θ − 1

1 + cot²θ = csc²θ

Divide first identity by sin²θ

Also: csc²θ − cot²θ = 1
cot²θ = csc²θ − 1

Deriving the Other Two Identities

Both derived identities follow the same two-step process: divide every term in sin²θ + cos²θ = 1 by the same expression, then simplify each ratio using the definitions of the trig functions.

Deriving tan²θ + 1 = sec²θ (divide by cos²θ)

Start with: sin²θ + cos²θ = 1
Divide every term by cos²θ: sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
Simplify each ratio: (sin θ/cos θ)² + 1 = (1/cos θ)²
Apply definitions (tan = sin/cos, sec = 1/cos): tan²θ + 1 = sec²θ ✓

Deriving 1 + cot²θ = csc²θ (divide by sin²θ)

Start with: sin²θ + cos²θ = 1
Divide every term by sin²θ: sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
Simplify each ratio: 1 + (cos θ/sin θ)² = (1/sin θ)²
Apply definitions (cot = cos/sin, csc = 1/sin): 1 + cot²θ = csc²θ ✓

Useful Rearrangements

Each identity has six equivalent forms depending on which term you isolate. Recognizing these rearrangements on sight is what makes simplification problems fast.

Form Equivalent rearrangements Most useful for
sin²θ + cos²θ = 1 sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ		 Replacing sin² or cos² when the other is known
tan²θ + 1 = sec²θ tan²θ = sec²θ − 1
sec²θ − tan²θ = 1		 Problems involving tan and sec together
1 + cot²θ = csc²θ cot²θ = csc²θ − 1
csc²θ − cot²θ = 1		 Problems involving cot and csc together

💡 Pattern Recognition Shortcut

When you see 1 − sin²θ in a problem, immediately replace it with cos²θ. When you see 1 − cos²θ, replace with sin²θ. When you see sec²θ − 1, replace with tan²θ. Training yourself to recognize these patterns on sight is the core skill for simplification and proof problems.

Substitution Map — If You See This, Replace With That

This is the table to have open during homework. Every row is a pattern that appears on exams — recognizing it on sight is faster than deriving it each time.

If you see… Replace with… Why / identity used
sin²θ + cos²θ 1 The gold standard — collapses immediately
1 − sin²θ cos²θ Rearrangement of identity 1 — solving for cos²
1 − cos²θ sin²θ Rearrangement of identity 1 — solving for sin²
tan²θ + 1 sec²θ Identity 2 — appears after factoring tan²
sec²θ − 1 tan²θ Rearrangement of identity 2 — very common in proofs
sec²θ − tan²θ 1 Identity 2 rearranged — collapses like the first identity
1 + cot²θ csc²θ Identity 3 — appears when cot and csc are both present
csc²θ − 1 cot²θ Rearrangement of identity 3
csc²θ − cot²θ 1 Identity 3 rearranged — collapses to 1
(1 − sin θ)(1 + sin θ) cos²θ Difference of squares → 1 − sin²θ = cos²θ
(1 − cos θ)(1 + cos θ) sin²θ Difference of squares → 1 − cos²θ = sin²θ

⚠️ The “Squares” Rule

If you see any trig function squared in a homework problem, a Pythagorean identity is almost certainly the intended tool. If you see the number 1 being added or subtracted from a squared trig function, an identity is definitely hiding there.

Finding Missing Trig Values

One of the most common uses of the Pythagorean identity: given one trig value and the quadrant, find the rest. The identity sin²θ + cos²θ = 1 lets you find either function when you know the other. Step 3 of this process — applying the quadrant sign — requires knowing which functions are positive where. The diagram below is the reference.

Quadrant diagram showing which trig functions are positive in each quadrant. All six are positive in Q1. Only sine and cosecant are positive in Q2. Only tangent and cotangent in Q3. Only cosine and secant in Q4. Memory device: All Students Take Calculus.

Reciprocals share signs — csc goes with sin, sec goes with cos, cot goes with tan. Only need to remember ASTC for the three primary functions.

1

Write the identity with the known value substituted

If sin θ = 3/5, substitute: (3/5)² + cos²θ = 1 → 9/25 + cos²θ = 1

2

Solve for the unknown function squared

cos²θ = 1 − 9/25 = 16/25 → cos θ = ±4/5

3

Apply the quadrant to determine the sign

If θ is in Q II, cos is negative → cos θ = −4/5. Then tan θ = sin θ/cos θ = (3/5)/(−4/5) = −3/4.

📐 Worked Example — All Six Functions From One

Given: tan θ = 5/12 and θ is in Q III. Find sin θ, cos θ, csc θ, sec θ, and cot θ.

Step 1: Use tan²θ + 1 = sec²θ: (5/12)² + 1 = sec²θ → 25/144 + 144/144 = 169/144 → sec θ = ±13/12.

Step 2: Q III: cos is negative, so sec is negative → sec θ = −13/12 → cos θ = −12/13.

Step 3: sin²θ = 1 − cos²θ = 1 − 144/169 = 25/169 → sin θ = ±5/13. Q III: sin is negative → sin θ = −5/13.

Results: sin θ = −5/13, cos θ = −12/13, csc θ = −13/5, sec θ = −13/12, cot θ = 12/5.

Simplifying Expressions

The most common exam task: an expression that looks complicated, but substituting a Pythagorean identity reduces it to something simple. The key skill is recognizing which substitution to make.

Example 1 — Direct substitution

Simplify: sin²θ + cos²θ + tan²θ

Recognize sin²θ + cos²θ = 1, so substitute: 1 + tan²θ

Recognize 1 + tan²θ = sec²θ. Answer: sec²θ.

Example 2 — Factoring pattern

Simplify: (1 − cos θ)(1 + cos θ)

This is a difference of squares: (1 − cos θ)(1 + cos θ) = 1 − cos²θ

Recognize 1 − cos²θ = sin²θ. Answer: sin²θ.

Example 3 — Substitution then simplify

Simplify: (sin²θ − 1)/cos θ

Recognize sin²θ − 1 = −(1 − sin²θ) = −cos²θ

Substitute: −cos²θ/cos θ = −cos θ. Answer: −cos θ.

Example 4 — Converting to sin and cos first

Simplify: sec²θ − tan²θ − sin²θ

Recognize sec²θ − tan²θ = 1 (rearrangement of identity 2)

Substitute: 1 − sin²θ = cos²θ. Answer: cos²θ.

Proving Identities

Proof problems ask you to show that two expressions are equal. The strategy is different from simplification — and most students get it wrong by trying to manipulate both sides simultaneously.

⚠️ The Cardinal Rule of Proving Identities

Work on one side only. Choose the more complex side and transform it until it matches the other side. Never move terms across the equals sign, and never assume what you’re trying to prove. Working both sides toward a middle point is not a valid proof — it assumes what you’re trying to show.

Practical strategy for most proof problems:

1

Choose the more complex side to work on.

2

Convert everything to sin and cos — this reduces the number of distinct functions and makes patterns visible.

3

Look for Pythagorean identity substitutions: 1−sin², 1−cos², sec²−1, csc²−1, etc.

4

Factor, combine fractions, or cancel as needed until the expression matches the other side.

Proof Example 1 — Prove: cos²θ/(1 − sin θ) = 1 + sin θ

Work the left side only. Replace cos²θ with 1 − sin²θ:

(1 − sin²θ)/(1 − sin θ)

Factor the numerator as a difference of squares: (1 − sin θ)(1 + sin θ)/(1 − sin θ)

Cancel the (1 − sin θ) factor: = 1 + sin θ ✓ Matches the right side.

Proof Example 2 — Prove: (1/sin θ) − (cos²θ/sin θ) = sin θ

This type uses a common denominator to combine fractions — one of the most common proof structures on Hawkes and WebAssign.

Work the left side only. Both terms already share the denominator sin θ, so combine:

(1 − cos²θ)/sin θ

Recognize 1 − cos²θ = sin²θ (Pythagorean identity):

sin²θ/sin θ

Cancel one factor of sin θ: = sin θ ✓ Matches the right side.

Common Mistakes

❌ Writing sin²θ + cos²θ = 1 as sinθ + cosθ = 1

The squares are non-negotiable. sin θ + cos θ is not equal to 1 in general — that would only be true at specific angles. The identity only holds with the squares.

❌ Writing the second identity as tan²θ + 1 = sec θ (missing the square on sec)

The correct form is tan²θ + 1 = sec²θ — both sides are squared. A common error under pressure is to write sec instead of sec². The derivation makes this obvious: 1/cos²θ = sec²θ, not sec θ.

❌ Forgetting the ± when taking the square root

When solving cos²θ = 16/25, cos θ = ±4/5. The sign is determined by the quadrant, not by the identity. Skipping the ± and writing cos θ = 4/5 directly will give the wrong answer whenever the angle is in Q II or Q III.

❌ Working both sides of a proof simultaneously

Manipulating both sides toward a common middle is not a valid proof strategy — it assumes the conclusion. Choose the more complex side and transform it alone until it equals the other side unchanged.

❌ Mixing up which identity to use with tan/cot

tan²θ + 1 = sec²θ involves tan and sec together. 1 + cot²θ = csc²θ involves cot and csc together. Never mix them — tan doesn’t appear in the csc identity, and cot doesn’t appear in the sec identity.

❌ Angle mismatch: sin²α + cos²β ≠ 1

The identity only holds when both functions use the same angle. sin²α + cos²α = 1 and sin²β + cos²β = 1 are both true, but sin²α + cos²β is just two unrelated numbers that don’t simplify to anything. This trap appears in multi-variable problems and in calculus when expressions involve both θ and 2θ — sin²(2θ) + cos²(2θ) = 1 is fine, but sin²θ + cos²(2θ) is not an identity.

❌ Confusing sin²θ with sin(2θ)

sin²θ means (sin θ)² — the sine value is squared. sin(2θ) means sine of the angle 2θ — a completely different value. At θ = 30°: sin²(30°) = (0.5)² = 0.25, but sin(2 × 30°) = sin(60°) = 0.866. These are not the same. The confusion is easy to make in handwritten work and on platforms that render both similarly. Always check whether the exponent is on the function or the angle.

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Practice Problems

📝 Problem 1 — Finding a Missing Value

Given cos θ = −5/13 and θ is in Q III, find sin θ, tan θ, and sec θ.

Click to reveal answer

sin²θ = 1 − cos²θ = 1 − 25/169 = 144/169 → sin θ = ±12/13. Q III: sin is negative → sin θ = −12/13

tan θ = sin θ/cos θ = (−12/13)/(−5/13) = 12/5 → tan θ = 12/5 (positive in Q III ✓)

sec θ = 1/cos θ = 1/(−5/13) = −13/5 → sec θ = −13/5

📝 Problem 2 — Simplification

Simplify: tan²θ · cos²θ + cos²θ

Click to reveal answer

Factor out cos²θ: cos²θ(tan²θ + 1)

Recognize tan²θ + 1 = sec²θ: cos²θ · sec²θ

Since sec θ = 1/cos θ, sec²θ = 1/cos²θ: cos²θ · (1/cos²θ) = 1

📝 Problem 3 — Proof

Prove: sin²θ · sec²θ = sec²θ − 1

Click to reveal answer

Work the left side only.

sin²θ · sec²θ = sin²θ · (1/cos²θ) = sin²θ/cos²θ = (sin θ/cos θ)² = tan²θ

Recognize tan²θ = sec²θ − 1 (rearrangement of second Pythagorean identity)

Left side = sec²θ − 1 = right side ✓

The Calculus Connection

Pythagorean identities appear throughout calculus — not just as a topic but as a constant tool for simplification. Students who can’t apply these identities quickly pay a significant time tax on every trig-heavy problem in Calc I and II.

Integration by trig substitution is the most intensive use. When an integrand contains √(1 − x²), the substitution x = sin θ converts it to √(1 − sin²θ) = √(cos²θ) = cos θ — a much simpler form. When an integrand contains √(1 + x²), the substitution x = tan θ converts it to √(1 + tan²θ) = √(sec²θ) = sec θ. Without the identities, this substitution goes nowhere.

Simplifying derivatives of trig functions frequently produces expressions like sin²θ + cos²θ that immediately collapse to 1. For example, differentiating sin θ · cos θ using the product rule gives cos²θ − sin²θ, which the double angle identity converts to cos(2θ) — but first recognizing when Pythagorean identities apply speeds every step of the process.

Power reduction is another major use. Integrals involving sin²θ or cos²θ (which appear constantly in Calc II) use the identities cos²θ = (1 + cos 2θ)/2 and sin²θ = (1 − cos 2θ)/2 — both derived directly from sin²θ + cos²θ = 1 combined with the double angle formulas.

💡 The Derivative of sec x — Pythagorean Identity in Action

The derivative of tan x is sec²x. This isn’t a coincidence — it comes directly from differentiating sin x/cos x using the quotient rule, which produces (cos²x + sin²x)/cos²x = 1/cos²x = sec²x. The Pythagorean identity is hidden inside the derivative formula itself.

Platform Tips

ALEKS — simplification problems on ALEKS often require a specific form for the answer. If your simplified answer involves sec²θ, try entering it as 1/cos²(θ) if the platform doesn’t recognize the sec notation directly. ALEKS proof-style problems typically ask you to identify which identity was applied at each step — knowing the names (Pythagorean identity, reciprocal identity, quotient identity) matters as much as knowing the algebra.

Hawkes Learning — Certify mode frequently presents “simplify the expression” problems where the intended path is a Pythagorean substitution followed by cancellation. If your answer isn’t matching, check whether you’ve fully simplified — Hawkes expects the most reduced form, so cos²θ is acceptable but sin²θ/cos²θ · cos²θ is not.

WebAssign — proof problems on WebAssign often use a fill-in-the-steps format where you select which identity justifies each line. Know all three identities by name and their rearrangements. The answer to “what justifies replacing 1 − cos²θ with sin²θ” is “Pythagorean Identity,” not “algebra.”

Knewton Alta — adaptive problems in trig identities will escalate to proof-style questions once you’ve demonstrated basic simplification. The platform weights these heavily. Working through the proof examples on this page before attempting Knewton Alta certify problems will save time.

MyMathLab — entering answers with exponents on trig functions requires using the exponent tool in the equation editor. Type sin, then use the x² button to raise it to the second power, rather than typing sin²θ as plain text. Plain text entry is often rejected even if mathematically correct.

Frequently Asked Questions

Do I need to memorize all three Pythagorean identities?

Only the first one — sin²θ + cos²θ = 1. The other two are derived by dividing by cos²θ (to get tan²θ + 1 = sec²θ) and by sin²θ (to get 1 + cot²θ = csc²θ). If you understand the derivation, you can reconstruct the others in about 30 seconds.

Why is it called the Pythagorean identity?

Because it’s the Pythagorean theorem (a² + b² = c²) applied to the unit circle. The legs of the right triangle inside the unit circle are cos θ and sin θ, and the hypotenuse is the radius = 1. Substituting gives cos²θ + sin²θ = 1. It’s not a separate theorem — it’s the same theorem in a different context.

Is sin²θ + cos²θ = 1 always true, even for obtuse angles?

Yes — for every angle without exception. The identity comes from the unit circle, not from right triangles. Even for 135°, 270°, or −400°, the point P = (cos θ, sin θ) is on the unit circle, and the unit circle always satisfies x² + y² = 1. There are no special cases or exceptions.

How do I know which identity to use when simplifying?

Match the functions in the problem to the functions in the identity. If the problem has sin and cos, use sin²θ + cos²θ = 1. If it has tan and sec, use tan²θ + 1 = sec²θ. If it has cot and csc, use 1 + cot²θ = csc²θ. When in doubt, convert everything to sin and cos first — that reduces the problem to using only the first identity.

What’s the difference between a trig identity and a trig equation?

An identity is true for all values of θ — sin²θ + cos²θ = 1 holds at every angle. An equation is true only at specific values — sin θ = 0.5 is only true at θ = 30°, 150°, and their coterminal angles. When you prove an identity, you’re showing it’s universally true. When you solve an equation, you’re finding the specific angles that make it true.

Can I use the Pythagorean identity to verify a specific angle?

Yes — and it’s a useful calculator check. After computing sin(θ) and cos(θ), square both and add them. The result should be exactly 1. If you get something like 0.9999 or 1.0001, it’s a rounding error from the calculator, not a violation of the identity. If you get something significantly different from 1, either the angle was entered incorrectly or you’re in the wrong mode.

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