Every identity family, a proven strategy flowchart, and 10 step-by-step worked proofs. This is the complete resource for verifying and proving trigonometric identities.
These identities define the six trig functions in terms of each other. They come directly from the definitions of sin, cos, tan, csc, sec, and cot on the unit circle. Memorize all six — they are the starting point for almost every identity proof.
| Identity | Alternate Form | Key Point |
|---|---|---|
| csc θ = 1 / sin θ | sin θ = 1 / csc θ | csc is the reciprocal of sin |
| sec θ = 1 / cos θ | cos θ = 1 / sec θ | sec is the reciprocal of cos |
| cot θ = 1 / tan θ | tan θ = 1 / cot θ | cot is the reciprocal of tan |
Tangent
tan θ = sin θ / cos θ
sin over cos — tangent is the slope of the terminal side
Cotangent
cot θ = cos θ / sin θ
cos over sin — the reciprocal relationship
Proof Tip: Convert Everything to Sin and Cos First
When you are stuck on a proof, replace every tan, cot, sec, and csc using the identities above. You will then have an expression involving only sin and cos, which is far easier to simplify algebraically.
All three Pythagorean identities come from the unit circle equation x squared plus y squared equals 1. Substituting cos theta for x and sin theta for y gives the first identity. Dividing by cos squared theta or sin squared theta produces the other two.
Form 1 — The Foundation
sin²θ + cos²θ = 1
From unit circle: x² + y² = 1
sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ
These two rearrangements appear constantly in proofs
Form 2 — Divide by cos²θ
1 + tan²θ = sec²θ
Divide Form 1 by cos²θ
tan²θ = sec²θ − 1
sec²θ − tan²θ = 1
Use when sec or tan appears in the expression
Form 3 — Divide by sin²θ
1 + cot²θ = csc²θ
Divide Form 1 by sin²θ
cot²θ = csc²θ − 1
csc²θ − cot²θ = 1
Use when csc or cot appears in the expression
Which Pythagorean Form to Use — Decision Guide
Negative angle identities tell you what happens when you replace theta with negative theta. The behavior depends on whether the function is even (symmetric about the y-axis) or odd (symmetric about the origin).
cos(−θ) = cos θ
sec(−θ) = sec θ
Graph is symmetric about the y-axis. Cosine is the only even basic trig function; secant inherits this because sec = 1/cos.
sin(−θ) = −sin θ
tan(−θ) = −tan θ
cot(−θ) = −cot θ
csc(−θ) = −csc θ
Graph is symmetric about the origin. Sin, tan, cot, and csc are all odd — they pick up a negative sign when the angle is negated.
Why Even and Odd?
On the unit circle, the point at angle theta is (cos theta, sin theta). The point at angle negative theta is (cos theta, negative sin theta) — same x-coordinate, opposite y-coordinate. This is exactly the definition of an even function for x-coordinates and an odd function for y-coordinates. Every other trig function follows from these two facts.
Deriving tan(−θ)
tan(−θ) = sin(−θ) / cos(−θ)
= (−sin θ) / cos θ
= −(sin θ / cos θ)
= −tan θ
Cofunction identities relate each trig function to its complement (90 degrees minus the angle). The word "cofunction" is where the "co-" prefix in cosine, cotangent, and cosecant comes from — they are the cofunctions of sine, tangent, and secant.
Degree Form (90°)
sin(90° − θ) = cos θ
cos(90° − θ) = sin θ
tan(90° − θ) = cot θ
cot(90° − θ) = tan θ
sec(90° − θ) = csc θ
csc(90° − θ) = sec θ
Radian Form (π/2)
sin(π/2 − θ) = cos θ
cos(π/2 − θ) = sin θ
tan(π/2 − θ) = cot θ
cot(π/2 − θ) = tan θ
sec(π/2 − θ) = csc θ
csc(π/2 − θ) = sec θ
Memory Pattern: Cofunctions Switch Pairs
The pattern is: sin pairs with cos, tan pairs with cot, sec pairs with csc. Any function of an angle equals the cofunction of the complement. In a right triangle where the two acute angles are theta and (90° minus theta), the sine of one equals the cosine of the other — that is literally the origin of the word "cosine."
Proving trig identities requires a different mindset than solving equations. There is no single algorithm that works every time, but there is a strategy flowchart that handles the vast majority of precalculus problems.
The Golden Rule
Work on ONE side only. Do not touch the other side.
You are not solving an equation — you are transforming one expression into another. Cross-multiplying, adding to both sides, or moving terms across the equal sign assumes the identity is true before you prove it. That is circular reasoning and earns zero credit.
Strategy Flowchart — Apply in Order
Choose the more complex side
Work on the side with more terms, fractions, or compound expressions. It is easier to simplify than to complicate. If both sides look equal in complexity, try the left side first.
Convert everything to sin and cos
Replace tan, cot, sec, and csc using reciprocal and quotient identities. You now have a single-function algebra problem. This step alone unlocks most proofs.
tan θ → sin θ / cos θ
sec θ → 1 / cos θ
csc θ → 1 / sin θ
cot θ → cos θ / sin θ
Apply Pythagorean substitution immediately
The moment you see sin²θ + cos²θ, 1 − sin²θ, 1 − cos²θ, 1 + tan²θ, or any other Pythagorean pattern, substitute right away. Do not wait.
Simplify fractions — combine or split
If you have a sum of fractions, get a common denominator and combine. If you have one fraction with a sum in the numerator, split it into separate fractions. Look for cancellation after every step.
Factor when you can
Look for difference of squares: sin²θ − cos²θ = (sin θ + cos θ)(sin θ − cos θ). Factor out common trig functions. Factor using quadratic patterns with sin² as the variable.
Multiply by the conjugate
When you see a binomial like (1 − sin θ), (1 + cos θ), or (sec θ − tan θ), multiply top and bottom by its conjugate. The result will always collapse to a Pythagorean expression.
(1 − sin θ)(1 + sin θ) = 1 − sin²θ = cos²θ
(sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1
If stuck, try the other side
If you are completely stuck on one side, start fresh on the other side. Work both sides toward a common middle expression. This is allowed as long as you show each side independently reaching that middle, without treating them as equal along the way.
Quick Decision Tree: What Do I See?
These proofs progress from basic (Proof 1) to challenging (Proof 10). Each one is worked step by step with commentary explaining why each move was made. Study the commentary — the reasoning is as important as the algebra.
Verify: tan θ · cos θ = sin θ
Strategy used: convert tan to sin/cos, cancel common factor.
Verify: sin²θ + cos²θ + tan²θ = sec²θ
Strategy used: recognize and apply two Pythagorean identities sequentially.
Verify: (sin θ / csc θ) + (cos θ / sec θ) = 1
Strategy used: reciprocal identity for csc and sec, then Pythagorean identity.
Verify: (1/sin θ) − (1/csc θ) = csc θ − sin θ
LHS check:
(1/sin θ) − (1/csc θ) = (1/sin θ) − sin θ
Both sides reach (1 − sin²θ)/sin θ = cos²θ/sin θ ✓
Strategy used: reciprocal substitution, common denominator, Pythagorean substitution.
Verify: sin θ / (1 − cos θ) = (1 + cos θ) / sin θ
Strategy used: multiply by conjugate, difference of squares, Pythagorean substitution, cancel.
Verify: (sin θ)/(1 + cos θ) + (1 + cos θ)/(sin θ) = 2 csc θ
Strategy used: common denominator, expand, Pythagorean substitution, factor, cancel.
Verify: (sec θ − tan θ)(sec θ + tan θ) = 1
Strategy used: recognize difference of squares, apply Pythagorean Form 2. Clean and fast — always look for this pattern with sec and tan.
Verify: (tan θ + sin θ) / (tan θ − sin θ) = (1 + cos θ) / (1 − cos θ)
Strategy used: convert tan, factor common sin θ, rewrite as compound fraction, simplify by dividing.
Verify: sin(−θ) / cos(−θ) = −tan θ
Strategy used: even/odd identities for negative angles, then quotient identity. This also proves tan is an odd function.
Verify: (sec θ + csc θ) / (sin θ + cos θ) = sec θ · csc θ
Strategy used: convert all functions to sin/cos, build common denominator in the nested numerator, cancel common binomial factor, recognize product of reciprocals.
These are the errors that cost points on exams. They each represent a logical breakdown — not just an algebra slip.
Mistake 1: Cross-Multiplying Both Sides
Cross-multiplying assumes the two sides are equal — but that is exactly what you are trying to prove. This is circular reasoning.
Trying to prove A/B = C/D by writing A·D = B·C ← INVALID
Instead: simplify A/B alone until it equals C/D
Mistake 2: Adding to Both Sides
You cannot add, subtract, multiply, or divide both sides during a verification. That is solving an equation. Treat each side as a separate expression to simplify.
Verifying A = B by adding C to both sides: A+C = B+C ← INVALID
Instead: transform A alone into B, without touching B
Mistake 3: Forgetting the Alternate Pythagorean Forms
Students memorize sin²θ + cos²θ = 1 but forget that 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ are identical rearrangements. The moment you see (1 − sin²θ) in a denominator, replace it with cos²θ — do not leave it.
1 − sin²θ = cos²θ
1 − cos²θ = sin²θ
sec²θ − 1 = tan²θ
csc²θ − 1 = cot²θ
Mistake 4: Splitting a Fraction Incorrectly
You can split a fraction when the denominator is a single term. You cannot split when the numerator is a single term over a sum.
Valid: (sin θ + cos θ)/sin θ = 1 + (cos θ/sin θ) = 1 + cot θ
Invalid: sin θ/(sin θ + cos θ) ≠ 1 + sin θ/cos θ
Mistake 5: Squaring Both Sides
Squaring can introduce extraneous solutions and is not a valid proof step. If you square both sides and get a true result, you have not proven the original identity — you have only shown the squares are equal (which happens when the originals are equal OR opposite in sign).
These identities appear in Chapter 7 of Stewart and are frequently used to compute exact values and to establish other identities. Learn the sign patterns: sine sum uses plus-plus and minus-minus; cosine sum uses plus-minus and minus-plus.
Sine
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
Cosine
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
Tangent
tan(A + B) = (tan A + tan B) / (1 − tan A tan B)
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Memory Aid: FOIL for Sine, Reverse FOIL for Cosine
For sine: the terms mix the functions (sin·cos + cos·sin). The sign in the answer matches the sign in the angle. For cosine: the terms keep matching functions (cos·cos − sin·sin). The sign in the answer is OPPOSITE to the sign in the angle.
Worked Example: Exact Value of cos(15°)
cos(15°) = cos(45° − 30°)
= cos 45° cos 30° + sin 45° sin 30°
= (√2/2)(√3/2) + (√2/2)(1/2)
= √6/4 + √2/4
= (√6 + √2) / 4
Double-angle formulas come directly from the sum formulas when you set A = B = theta. Cosine gives three equivalent forms — you need all three because each is useful in different proof contexts.
sin(2θ)
sin(2θ) = 2 sin θ cos θ
One form only — easy to remember
tan(2θ)
tan(2θ) = 2 tan θ / (1 − tan²θ)
Derived from sin(2θ)/cos(2θ)
cos(2θ) — Three Forms
cos(2θ) = cos²θ − sin²θ
Direct from cos(A+B) with A = B = θ
cos(2θ) = 2cos²θ − 1
Replace sin²θ with 1 − cos²θ in the first form
cos(2θ) = 1 − 2sin²θ
Replace cos²θ with 1 − sin²θ in the first form
Which cos(2θ) Form to Use
Use cos²θ − sin²θ when you want to keep both functions present.
Use 2cos²θ − 1 when you want to eliminate sin (useful if the other side has only cos).
Use 1 − 2sin²θ when you want to eliminate cos (useful if the other side has only sin).
Power-Reducing Formulas (Derived from Double-Angle)
sin²θ = (1 − cos 2θ) / 2
cos²θ = (1 + cos 2θ) / 2
These come from solving cos(2θ) = 1 − 2sin²θ and cos(2θ) = 2cos²θ − 1 for sin²θ and cos²θ respectively. Essential for calculus integration.
This distinction is tested directly on exams and is essential for understanding what you are doing in each problem type.
Trigonometric Identity
True for ALL values of the variable (where both sides are defined)
Cannot be solved — it is proved (verified)
Used to simplify, rewrite, or transform expressions
Has infinitely many solutions — every valid angle
sin²θ + cos²θ = 1
True for every value of θ
Trigonometric Equation
True only for specific values of the variable
Is solved to find those specific values
Produces a solution set, often with periodic repeating pattern
Has finitely many solutions in any given interval
sin θ = 1/2
True only for θ = π/6 + 2kπ and θ = 5π/6 + 2kπ
Practical Test: Is It an Identity or an Equation?
Identities Are Used in Both Directions
When proving identities, you always work from one side to the other. But when solving equations, you can use any known identity to rewrite both sides freely. For example, when solving cos(2θ) = cos θ, you choose the form of cos(2θ) that makes the equation easiest to factor — you are not constrained to work on one side only.
Work on ONE side only. Never move terms from one side of the equal sign to the other — that assumes the identity is true, which is circular reasoning. Pick the more complex side and simplify it until it matches the other side exactly.
The three Pythagorean identities are: (1) sin squared theta plus cos squared theta equals 1, derived from the unit circle equation x squared plus y squared equals 1; (2) 1 plus tan squared theta equals sec squared theta, derived by dividing the first identity by cos squared theta; (3) 1 plus cot squared theta equals csc squared theta, derived by dividing the first identity by sin squared theta. Each identity has two alternate rearrangements useful in proofs.
Cosine and secant are even functions: cos(-theta) = cos(theta) and sec(-theta) = sec(theta). Sine, tangent, cotangent, and cosecant are odd functions: sin(-theta) = -sin(theta), tan(-theta) = -tan(theta), cot(-theta) = -cot(theta), csc(-theta) = -csc(theta). Even means the graph is symmetric about the y-axis; odd means symmetric about the origin.
The two most common mistakes are: (1) Cross-multiplying both sides or adding terms to both sides, which assumes the identity is already true — this is circular and invalid. (2) Changing both sides simultaneously. You must pick one side and transform it alone. A secondary mistake is forgetting to check alternate forms of Pythagorean identities: 1 minus sin squared theta equals cos squared theta, and 1 minus cos squared theta equals sin squared theta.
Multiply by the conjugate when you see a binomial expression like (1 plus sin theta), (1 minus cos theta), (sec theta plus tan theta), or similar forms in a denominator or being divided. Multiplying by the conjugate over itself creates a difference-of-squares pattern that often reduces to a Pythagorean identity. For example, (1 minus sin theta)(1 plus sin theta) equals 1 minus sin squared theta equals cos squared theta.
A trig identity is true for ALL values of the variable (wherever both sides are defined). A trig equation is true only for specific values. For example, sin squared theta plus cos squared theta equals 1 is an identity — it holds for every theta. But sin theta equals 0.5 is an equation with solutions theta equals pi/6 plus 2k-pi and theta equals 5pi/6 plus 2k-pi. Identities are used to simplify; equations are solved for specific solutions.
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