Reference

Identity Definition and Examples

Concept

Identity

An identity is an equation that is always true. It may or may not contain variables. If an identity contains variables, no matter what value the variables take, the equation will always hold true.

Identity Without Variables 3 = 3 Identity With Variables (a - b) (a + b) = a^{2} - b^{2}

Rule

Identity Property of Addition

Adding $0$ to any number always results in the number itself.

a + 0 = a

Because of this, $0$ is called the Additive Identity.

Proof

Informal Justification

Consider a number

a .

By the Reflexive Property of Equality,

a

is equal to itself.

a = a

Let

b

be another number. If

b

is added to and subtracted from the left-hand side of the above equation, the equality still holds true.

a = a \Leftrightarrow a + b - b = a

Finally,

b - b

is equal to

0 .

a + b - b = a \Leftrightarrow a + 0 = a ✓

It has been shown that

a + 0 = a .

By the Commutative Property of Addition,

0 + a

is also equal to

a .

Rule

Identity Property of Multiplication

Any number multiplied by $1$ is equal to the number itself.

a \cdot 1 = a

Because of this, the number $1$ is called the Multiplicative Identity.

Proof

Informal Justification

Consider a number

a .

By the definition of multiplication,

a

multiplied by another number

n

can be written as

n

times the addition of

a .

a \cdot n = n times a + a + \dots + a

n = 1,

the sum has only one term.

a \cdot 1 = 1 time a

Therefore,

a \cdot 1

is equal to

a .

Also, by the Commutative Property of Multiplication,

1 \cdot a = a .

Rule

Pythagorean Identities

For any angle $θ,$ the following trigonometric identities hold true.

sin^{2} θ + cos^{2} θ = 1

1 + tan^{2} θ = sec^{2} θ

1 + cot^{2} θ = csc^{2} θ

Proof

For Acute Angles

Consider a right triangle with a hypotenuse of

1 .

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to

∠ θ

can be expressed in terms of the angle.

	Definition	Substitute	Simplify
$sin θ$	$\frac{Length of opposite side to ∠ θ}{Hypotenuse}$	$\frac{opp}{1}$	$opp$
$cos θ$	$\frac{Length of adjacent side to ∠ θ}{Hypotenuse}$	$\frac{adj}{1}$	$adj$

It can be seen that if the hypotenuse of a right triangle is $1,$ the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.

By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of $s i n θ$ and $c o s θ$ is equal to the square of $1 .$

s i n^{2} θ + c o s^{2} θ ⇓ sin^{2} θ + cos^{2} θ = 1^{2} = 1

Since

cos θ

represents a side length, it is not

0 .

Therefore, by diving both sides of the above equation by

cos^{2} θ,

the second identity can be obtained.

sin^{2} θ + cos^{2} θ = 1

DivEqn

$LHS / cos^{2} θ = RHS / cos^{2} θ$

\frac{sin ^{2} θ + cos ^{2} θ}{cos ^{2} θ} = \frac{1}{cos ^{2} θ}

▼

Simplify

WriteSumFrac

Write as a sum of fractions

\frac{sin ^{2} θ}{cos ^{2} θ} + \frac{cos ^{2} θ}{cos ^{2} θ} = \frac{1}{cos ^{2} θ}

QuotOne

$\frac{a}{a} = 1$

\frac{sin ^{2} θ}{cos ^{2} θ} + 1 = \frac{1}{cos ^{2} θ}

WritePow

Write as a power

\frac{sin ^{2} θ}{cos ^{2} θ} + 1 = \frac{1 ^{2}}{cos ^{2} θ}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

(\frac{sin θ}{cos θ})^{2} + 1 = (\frac{1}{cos θ})^{2}

$\frac{sin θ}{cos θ} = tan θ$

tan^{2} θ + 1 = (\frac{1}{cos θ})^{2}

$\frac{1}{cos θ} = sec θ$

tan^{2} θ + 1 = sec^{2} θ

CommutativePropAdd

Commutative Property of Addition

1 + tan^{2} θ = sec^{2} θ

The second identity was obtained.

1 + tan^{2} θ = sec^{2} θ

Since

sin θ

represents a side length, it is not

0 .

Therefore, by dividing both sides of

sin^{2} θ + cos^{2} θ = 1

sin^{2} θ,

the third identity can be proven.

sin^{2} θ + cos^{2} θ = 1

DivEqn

$LHS / sin^{2} θ = RHS / sin^{2} θ$

\frac{sin ^{2} θ + cos ^{2} θ}{sin ^{2} θ} = \frac{1}{sin ^{2} θ}

▼

Simplify

WriteSumFrac

Write as a sum of fractions

\frac{sin ^{2} θ}{sin ^{2} θ} + \frac{cos ^{2} θ}{sin ^{2} θ} = \frac{1}{sin ^{2} θ}

QuotOne

$\frac{a}{a} = 1$

1 + \frac{cos ^{2} θ}{sin ^{2} θ} = \frac{1}{sin ^{2} θ}

WritePow

Write as a power

1 + \frac{cos ^{2} θ}{sin ^{2} θ} = \frac{1 ^{2}}{sin ^{2} θ}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

1 + (\frac{cos θ}{sin θ})^{2} = (\frac{1}{sin θ})^{2}

$\frac{cos θ}{sin θ} = cot θ$

1 + cot^{2} θ = (\frac{1}{sin θ})^{2}

$\frac{1}{sin θ} = csc θ$

1 + cot^{2} θ = csc^{2} θ

Finally, the third identity was obtained.

$1 + cot^{2} θ = csc^{2} θ$

Proof

For Any Angle

The first identity can be shown using the unit circle and the Pythagorean Theorem. Consider a point $(x, y)$ on the unit circle in the first quadrant, corresponding to the angle $θ .$ A right triangle can be constructed with $θ .$

By the Pythagorean Theorem, the sum of the squares of

x

and

y

equals

1 .

x^{2} + y^{2} = 1

In fact, this is true not only for points in the first quadrant, but for every point on the unit circle. Recall that, for points

(x, y)

on the unit circle corresponding to angle

θ,

it is known that

x = cos θ

and that

y = sin θ .

By substituting these expressions into the equation, the first identity can be obtained.

cos^{2} θ + sin^{2} θ = 1

Dividing both sides by either

cos^{2} θ

sin^{2} θ

leads to two variations of the Pythagorean Identity.

Rule

Square of a Binomial

When a binomial is squared, the resulting expression is a perfect square trinomial.

$(a + b)^{2} = a^{2} + 2 a b + b^{2} (a - b)^{2} = a^{2} - 2 a b + b^{2}$

For simplicity, depending on the sign of the binomial, these two identities can be expressed as one.

$(a \pm b)^{2} = a^{2} \pm 2 a b + b^{2}$

Proof

This rule will be first proven for $(a + b)^{2}$ and then for $(a - b)^{2} .$

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

This identity can be shown by first rewriting the square as a product.

(a + b)^{2}

PowToProdTwoFac

$a^{2} = a \cdot a$

(a + b) (a + b)

▼

Multiply parentheses

Distr

Distribute $(a + b)$

a (a + b) + b (a + b)

Distr

Distribute $a$

a^{2} + a b + b (a + b)

Distr

Distribute $b$

a^{2} + a b + b a + b^{2}

CommutativePropMult

Commutative Property of Multiplication

a^{2} + a b + a b + b^{2}

AddTerms

Add terms

a^{2} + 2 a b + b^{2}

It has been shown that

(a + b)^{2} = a^{2} + 2 a b + b^{2} .

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

In this case, when one term of the binomial is subtracted from the other, the middle term of the perfect square trinomial will instead be negative.

(a + b)^{2}

PowToProdTwoFac

$a^{2} = a \cdot a$

(a - b) (a - b)

▼

Multiply parentheses

Distr

Distribute $(a - b)$

a (a - b) - b (a - b)

Distr

Distribute $a$

a^{2} - a b - b (a - b)

Distr

Distribute $- b$

a^{2} - a b - b a + b^{2}

CommutativePropMult

Commutative Property of Multiplication

a^{2} - a b - a b + b^{2}

SubTerms

Subtract terms

a^{2} - 2 a b + b^{2}

It has been shown that

(a - b)^{2} = a^{2} - 2 a b + b^{2} .

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Proof

Proof

Proof

Proof

Proof

(a+b)2=a2+2ab+b2

(a−b)2=a2−2ab+b2

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$