Look for this. An identity claims equality throughout its shared domain. Look for a one-sided rewrite that exposes a familiar formula and keeps the original restrictions.
Record where the original expressions exist. Change one side using named identities or legal algebra, showing each equality until you reach the other side. A plotted match or one substituted angle can check a step but cannot establish the identity.
Worked example
Prove (1 − cos 2θ)/sin 2θ = tan θ and state the original restrictions in radians.
The left side needs sin 2θ ≠ 0, so θ ≠ kπ/2 for integer k. This also keeps tan θ defined. By the double-angle identities the left side becomes 2 sin²θ/(2 sin θ cos θ). Cancel 2 sin θ, which is nonzero on this domain, to get sin θ/cos θ = tan θ.
Try it yourself
Prove [sin(A + B) − sin(A − B)]/(2 cos A) = sin B. Identify the excluded values and explain which terms disappear.
Check your reasoning
Require cos A ≠ 0, so A ≠ π/2 + kπ for integer k. B is unrestricted.
Expand the numerator to (sin A cos B + cos A sin B) − (sin A cos B − cos A sin B). Subtracting cancels the two sin A cos B terms and leaves 2 cos A sin B.
Divide by 2 cos A. This factor is nonzero on the original domain, so the quotient is sin B.
Worked notes and diagrams
1
The words, first
The idea: the difference between simplify, verify and prove decides what a full answer looks like.
The first Pythagorean identity is Pythagoras on the unit circle: the sides are cos θ and sin θ, and the hypotenuse is the radius, 1.
Word
What it means
Identity
An equation true for every value where both sides exist.
Equation
A statement that two expressions are equal. Solving finds which allowed values make it true.
Simplify
Rewrite one expression more compactly. Nothing is being proved.
Verify
Show a statement holds for one particular value, usually by substituting.
Prove
Show it holds for every permitted value, by transforming one side into the other.
Restriction
A value excluded because it would make a denominator zero or a function undefined.
Reciprocal identity
csc x = 1/sin x for sin x ≠ 0 and sec x = 1/cos x for cos x ≠ 0. The form cot x = 1/tan x requires both sin x and cos x to be nonzero.
Quotient identity
tan θ = sin θ/cos θ when cos θ ≠ 0. cot θ = cos θ/sin θ when sin θ ≠ 0.
Pythagorean identity
sin²θ + cos²θ = 1, and the two that follow from dividing it through.
Conjugate
The same expression with the middle sign flipped — multiplying by it creates a difference of squares.
2
Reciprocal and quotient identities
The idea: every trig function can be written using just sin x and cos x. This is especially helpful when reciprocal factors can cancel.
Function
In sine and cosine
tan x
sin x / cos x
cot x
cos x / sin x
sec x
1 / cos x
csc x
1 / sin x
Worked example
Simplify csc x · tan x.
(1/sin x) · (sin x/cos x) = 1/cos x = sec x, on the original domain sin x ≠ 0 and cos x ≠ 0.
Watch out: sec goes with cos and csc goes with sin — the opposite of what the letters suggest. Each reciprocal pairs with the function whose name starts differently.
3
Pythagorean identities
The idea: on the unit circle, sin θ and cos θ are the legs of a right triangle with hypotenuse 1, so sin²θ + cos²θ = 1. The other two come from dividing that by cos²θ or sin²θ.
The Pythagorean theorem on the unit circle. It works for every angle, which is what makes it an identity.
Identity
Rearranged forms you will use
sin²x + cos²x = 1
1 − sin²x = cos²x
1 − cos²x = sin²x
1 + tan²x = sec²x
sec²x − 1 = tan²x
1 + cot²x = csc²x
csc²x − 1 = cot²x
Spot the pattern and substitute: (1 − sin²x)/cos x = cos²x/cos x = cos x for cos x ≠ 0. For (sec²x − 1)/tan x, require both cos x ≠ 0 and sin x ≠ 0. Then tan²x/tan x = tan x.
4
Sum, difference and double-angle identities
The idea: these let you find exact values for angles you do not know, like 75° or 105°, by building them from angles you do know.
sin(A ± B) = sin A cos B ± cos A sin B
cos(A ± B) = cos A cos B ∓ sin A sin B (note the sign flips)
tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)
sin 2A = 2 sin A cos A
cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A
Exact values
cos 105° = cos(60° + 45°) = cos 60° cos 45° − sin 60° sin 45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4. It comes out negative, which makes sense for Quadrant II.
Recognize the pattern backwards
2 sin 15° cos 15° is sin 2(15°) = sin 30° = 1/2. cos²(π/8) − sin²(π/8) is cos(π/4) = √2/2. Spotting these saves all the arithmetic.
Pick the form of cos 2A that fits
If you only know cos x = 1/3, use the version with only cosine: cos 2x = 2(1/3)² − 1 = −7/9.
Watch out: sin 2x is not 2 sin x, and sin(A + B) is not sin A + sin B. Functions do not distribute.
5
Proving identities and solving equations
The idea: to prove, work each side separately until they match. To solve, use an identity to get one function, then factor.
Proving: the playbook
Start with the more complicated side.
Pick a rewrite suggested by the expression. With reciprocal ratios, try sine and cosine.
Use a Pythagorean identity, find a common denominator, or factor.
Never move terms across the equals sign — you have not shown it is an equation yet.
Checking one value, like x = π/4, only shows the equation holds for that value. It is a useful sanity check, never a proof.
Non-permissible values
An identity only holds where both sides are defined. For sin x/(1 − cos x), the denominator is zero when cos x = 1, so x ≠ 2πn. Tangent and secant are undefined wherever cos x = 0.
Solving with identities
Solve sin 2x = sin x for 0 ≤ x < 2π:
Double-angle identity: 2 sin x cos x = sin x.
Move everything to one side and factor: sin x(2cos x − 1) = 0.
sin x = 0 gives 0 and π; cos x = 1/2 gives π/3 and 5π/3.
Four solutions: 0, π/3, π, 5π/3.
Watch out: dividing both sides by sin x instead of factoring. It throws away the solutions where sin x = 0.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Reciprocal and quotient identities · 4 questions
Practice 1
Which expression is equal to sec x?
1/tan x
cos x/sin x
1/cos x
1/sin x
Answer and explanation
1/cos x
Secant is the reciprocal of cosine. A quick way to remember: each reciprocal pairs with the function whose name starts with a different letter — sec with cos, csc with sin.
Practice 2
Which expression is equal to tan x?
sin x · cos x
sin x/cos x
1/sin x
cos x/sin x
Answer and explanation
sin x/cos x
tan x = y/x on the unit circle, and y = sin x while x = cos x. cos x/sin x is cot x.
Practice 3
Simplify sec x · cos x.
tan x
cos²x
1
sec²x
Answer and explanation
1
The original secant requires cos x ≠ 0. For these values (1/cos x) × cos x = 1.
Practice 4
Simplify csc x · tan x.
sin x
sec x
1
cot x
Answer and explanation
sec x
Write everything in sine and cosine: (1/sin x) · (sin x/cos x) = 1/cos x = sec x. The original expression requires both sin x ≠ 0 and cos x ≠ 0.
Pythagorean identities · 5 questions
Practice 1
Simplify (1 − cos²x) ÷ (sin x cos x).
sin x
1
tan x
cot x
Answer and explanation
tan x
The Pythagorean identity turns 1 − cos²x into sin²x. Then sin²x ÷ (sin x cos x) = sin x ÷ cos x = tan x. The original expression requires both sin x ≠ 0 and cos x ≠ 0.
Practice 2
sin²x + cos²x equals:
0
1
tan²x
sec²x
Answer and explanation
1
On the unit circle, sin x and cos x are the legs of a right triangle with hypotenuse 1, so the Pythagorean theorem gives sin²x + cos²x = 1.
Practice 3
Which of these is a true identity?
1 + tan²x = csc²x
1 + tan²x = sec²x
1 − tan²x = sec²x
tan²x − 1 = sec²x
Answer and explanation
1 + tan²x = sec²x
Divide sin²x + cos²x = 1 by cos²x and you get tan²x + 1 = sec²x. Dividing by sin²x instead gives 1 + cot²x = csc²x. Division by cos²x requires cos x ≠ 0, and division by sin²x requires sin x ≠ 0.
Practice 4
Simplify (1 − sin²x) ÷ cos x.
sec x
sin x
1
cos x
Answer and explanation
cos x
1 − sin²x = cos²x from the Pythagorean identity, and cos²x ÷ cos x = cos x. Keep the original restriction cos x ≠ 0.
Practice 5
Simplify (sec²x − 1) ÷ tan x.
tan x
1
sec x
cot x
Answer and explanation
tan x
Rearrange 1 + tan²x = sec²x to get sec²x − 1 = tan²x. Then tan²x ÷ tan x = tan x. The original expression requires both sin x ≠ 0 and cos x ≠ 0.
Sum, difference and double-angle · 7 questions
Practice 1
Using a sum identity, what is the exact value of sin 75°?
(√6 − √2)/4
(√6 + √2)/4
√2/2 + 1/2
(√3 + 1)/2
Answer and explanation
(√6 + √2)/4
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4. The minus version is sin 15°.
Practice 2
Which expression is equal to sin 2x?
sin²x − cos²x
sin x + cos x
2 sin x
2 sin x cos x
Answer and explanation
2 sin x cos x
The double-angle identity is sin 2x = 2 sin x cos x. Doubling the angle does not double the sine — 2 sin x is a different thing.
Practice 3
What is the exact value of 2 sin 15° cos 15°?
√3/2
1/4
1/2
√2/2
Answer and explanation
1/2
Recognize the double-angle pattern: 2 sin 15° cos 15° = sin 30° = 1/2. Spotting the identity saves any calculation.
Practice 4
What is the exact value of cos²(π/8) − sin²(π/8)?
√2/2
√3/2
1
1/2
Answer and explanation
√2/2
cos²A − sin²A = cos 2A, so this is cos(π/4) = √2/2. Do not confuse it with cos²A + sin²A, which is 1.
Practice 5
If cos x = 1/3, what is cos 2x?
2/3
1/9
7/9
−7/9
Answer and explanation
−7/9
Use the form of cos 2x written only in cosine: 2cos²x − 1 = 2(1/9) − 1 = −7/9. 2/3 wrongly doubles the cosine.
Practice 6
sin A = 3/5 and cos B = 5/13, with A and B both in Quadrant I. What is sin(A + B)?
−16/65
63/65
56/65
33/65
Answer and explanation
63/65
From 3-4-5 and 5-12-13 triangles, cos A = 4/5 and sin B = 12/13. Then sin(A + B) = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65.
Practice 7
Using a sum identity, what is the exact value of cos 105°?
(√2 + √6)/4
(√6 − √2)/4
−1/2
(√2 − √6)/4
Answer and explanation
(√2 − √6)/4
cos(60° + 45°) = cos 60° cos 45° − sin 60° sin 45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4. It is negative, as it must be in Quadrant II.
Proving and solving with identities · 4 questions
Practice 1
When proving a trigonometric identity, you should:
move terms from one side to the other, as when solving an equation
substitute one value of x and check both sides
square both sides first
work on each side separately until both sides match
Answer and explanation
work on each side separately until both sides match
An identity is what you are trying to prove, so you cannot treat it as a true equation yet. Simplify each side on its own until they are identical.
Practice 2
What is the non-permissible value of sin x ÷ (1 − cos x) for 0 ≤ x < 2π?
x = π/2 and x = 3π/2
x = 0
x = π
there is none
Answer and explanation
x = 0
The denominator is zero when cos x = 1, which happens at x = 0 in this domain. In general, x ≠ 2πn.
Practice 3
Solve sin 2x = sin x for 0 ≤ x < 2π.
π/3, 5π/3
0, π
0, π/3, π, 5π/3
π/6, 5π/6
Answer and explanation
0, π/3, π, 5π/3
Rewrite: 2 sin x cos x − sin x = 0, so sin x(2cos x − 1) = 0. sin x = 0 gives 0 and π; cos x = 1/2 gives π/3 and 5π/3. Dividing by sin x would lose half the answers.
Practice 4
Substituting x = π/4 makes both sides of an equation equal. What does this show?
That x = π/4 is the only solution the equation has on the interval
Only that the equation is true at x = π/4 — it does not prove an identity
That the equation is an identity, since both sides agree at x = π/4
That x = π/4 is a non-permissible value and must be excluded
Answer and explanation
Only that the equation is true at x = π/4 — it does not prove an identity
An identity must hold for every permissible value, so one matching value proves nothing on its own — two expressions can agree at a point and differ everywhere else. Only an algebraic proof shows it is always true. A single substitution also says nothing about how many solutions there are, and a value that makes both sides equal is by definition permissible, not excluded.
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