Reference

Trigonometric Identities

Rule

Reciprocal Identities

The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively.

$csc θ = \frac{1}{sin θ}$

$sec θ = \frac{1}{cos θ}$

$cot θ = \frac{1}{tan θ}$

Proof

Consider a right triangle with the three sides labeled with respect to an acute angle $θ .$

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written.

sin θ = \frac{opp}{hyp} cos θ = \frac{adj}{hyp} tan θ = \frac{opp}{adj} csc θ = \frac{hyp}{opp} sec θ = \frac{hyp}{adj} cot θ = \frac{adj}{opp}

The reciprocal of the sine ratio will now be calculated.

sin θ = \frac{opp}{hyp}

▼

Solve for

\frac{1}{sin θ}

MultEqn

$LHS \cdot \frac{1}{sin θ} = RHS \cdot \frac{1}{sin θ}$

1 = \frac{opp}{hyp} (\frac{1}{sin θ})

MultEqn

$LHS \cdot \frac{hyp}{opp} = RHS \cdot \frac{hyp}{opp}$

\frac{hyp}{opp} = \frac{1}{sin θ}

RearrangeEqn

Rearrange equation

\frac{1}{sin θ} = \frac{hyp}{opp}

It has been found that

\frac{1}{s i n θ},

which is the reciprocal of

sin θ,

is equal to

\frac{hyp}{opp} .

By the definition, the cosecant of

θ

is also the ratio of the lengths of the hypotenuse and the opposite side to

∠ θ .

Therefore, by the Transitive Property of Equality,

\frac{1}{s i n θ}

is equal to

csc θ .

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{1}{sin θ} = \frac{hyp}{opp} . csc θ = \frac{hyp}{opp} ⇓ csc θ = \frac{1}{sin θ}

By following a similar procedure, the other two identities for secant and cotangent can be proven.

Rule

Tangent and Cotangent Identities

The tangent of an angle $θ$ can be expressed as the ratio of the sine of $θ$ to the cosine of $θ .$

$tan θ = \frac{sin θ}{cos θ}$

Similarly, the cotangent of $θ$ can be expressed as the ratio of the cosine of $θ$ to the sine of $θ .$

$cot θ = \frac{cos θ}{sin θ}$

Proof

Tangent Identity

Two proofs will be written for this identity, one using a right triangle and the other using a unit circle.

Right Triangle

In a right triangle, the tangent of an angle $θ$ is defined as the ratio of the length of the opposite side $k$ to the length of the adjacent side $ℓ .$

At the same time, the sine and cosine of

θ

can be written as follows.

sin θ = \frac{k}{m} cos θ = \frac{ℓ}{m}

By manipulating the right-hand side of the equation

tan θ = \frac{k}{ℓ},

the tangent can be expressed as the sine over the cosine of

θ .

tan θ = \frac{k}{ℓ}

ReduceFrac

$\frac{a}{b} = \frac{a / m}{b / m}$

tan θ = \frac{k / m}{ℓ / m}

SubstituteII

$k / m = s i n (θ)$ , $ℓ / m = c o s (θ)$

tan θ = \frac{s i n θ}{c o s θ}

The proof of the identity is complete.

Unit Circle

Consider a unit circle and an angle $θ$ in standard position.

It is known that the point of intersection $P$ of the terminal side of the angle and the unit circle has coordinates $(cos θ, sin θ) .$

Draw a right triangle using the origin and $P (cos θ, sin θ)$ as two of its vertices. The length of the hypotenuse is $1$ and the lengths of the legs are $sin θ$ and $cos θ .$

As shown previously, the tangent of a right triangle is defined as the ratio of the length of the opposite side — in this case,

sin θ

— to the length of the adjacent side, which here is

cos θ .

tan θ = \frac{sin θ}{cos θ} ✓

Proof

Cotangent Identity

Two more proofs will be written for this identity, one of them using just a right triangle and the other using a unit circle.

Right Triangle

In a right triangle, the cotangent of an angle $θ$ is defined as the ratio of the length of the adjacent side $ℓ$ to the length of the opposite side $k .$

Additionally, the sine and cosine of

θ

can be written as follows.

sin θ = \frac{k}{m} cos θ = \frac{ℓ}{m}

By manipulating the right-hand side of the equation

cot θ = \frac{ℓ}{k},

the cotangent can be expressed as the cosine over the sine of

θ .

cot θ = \frac{ℓ}{k}

ReduceFrac

$\frac{a}{b} = \frac{a / m}{b / m}$

cot θ = \frac{ℓ / m}{k / m}

SubstituteII

$ℓ / m = c o s (θ)$ , $k / m = s i n (θ)$

cot θ = \frac{c o s θ}{s i n θ}

This proof is complete.

Unit Circle

Using a unit circle, it has been already proven that the tangent of an angle

θ

is the ratio of the sine to the cosine of the angle

θ .

tan θ = \frac{sin θ}{cos θ}

By manipulating the above equation, it can be shown that the cotangent of

θ

is the ratio of the cosine of

θ

to the sine of

θ .

tan θ = \frac{sin θ}{cos θ}

MultEqn

$LHS \cdot \frac{cos θ}{sin θ} = RHS \cdot \frac{cos θ}{sin θ}$

tan θ (\frac{cos θ}{sin θ}) = 1

DivEqn

$LHS / tan θ = RHS / tan θ$

\frac{cos θ}{sin θ} = \frac{1}{tan θ}

$cot (θ) = \frac{1}{tan ( θ )}$

\frac{cos θ}{sin θ} = cot θ

RearrangeEqn

Rearrange equation

cot θ = \frac{cos θ}{sin θ} ✓

This proof is complete.

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

Negative Angle Identities

The function $y = sin (x)$ has odd symmetry, and $y = cos (x)$ has even symmetry, which can be seen from their graphs. As a result, the corresponding identities hold true.

$sin (- θ) = - sin θ cos (- θ) = cos θ tan (- θ) = - tan θ$

Proof

sin (- θ) = - sin θ

and

cos (- θ) = cos θ

The identities will be proven using a unit circle. Consider an arbitrary $∠ θ$ on a unit circle. Let $A$ be the point that the angle forms on the circle.

Recall that the values on the $x -$ axis are represented by cosine and the value on the $y -$ axis are represented by sine. Therefore, the coordinates of $A$ are $(cos θ, sin θ) .$

Next, the point

A

can be reflected over the

x -

axis.

Since

A^{'}

is the same distance from the

y -

axis as

A,

its

x -

coordinate is also

cos θ .

Note that a reflection is a congruent transformation, so

A

and

A^{'}

are equidistant from the

x -

axis too. However,

A^{'}

is below the

x -

axis, so its

y -

coordinate is a negative

sin θ .

Additionally, after the reflection, the angle between $A^{'},$ the origin, and the $x -$ axis is $- θ .$ This means that the lengths of the newly created horizontal and vertical segments are $cos (- θ)$ and $sin (- θ) .$ Therefore, the coordinates of $A^{'}$ can also be written as $(cos (- θ), sin (- θ)) .$

This way there are two pairs of coordinates of the same point

A^{'} .

This allows to conclude the following.

A^{'} (c o s (- θ), s i n (- θ)) c o s (- θ) s i n (- θ) and A^{'} (c o s θ, - s i n θ) ⇓ = c o s θ = - s i n θ

Proof

tan (- θ) = - tan θ

By expressing

tan (- θ)

using sine and cosine, this identity can be shown.

tan (- θ)

$tan (θ) = \frac{sin ( θ )}{cos ( θ )}$

\frac{sin ( - θ )}{cos ( - θ )}

$sin (- θ) = - sin (θ)$

\frac{- sin ( θ )}{cos ( - θ )}

$cos (- θ) = cos (θ)$

\frac{- sin ( θ )}{cos ( θ )}

MoveNegNumToFrac

Put minus sign in front of fraction

- \frac{sin ( θ )}{cos ( θ )}

$\frac{sin ( θ )}{cos ( θ )} = tan (θ)$

- tan (θ)

Rule

Cofunction Identities

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

$sin (\frac{π}{2} - θ) = cos θ$

$cos (\frac{π}{2} - θ) = sin θ$

$tan (\frac{π}{2} - θ) = cot θ$

Proof

For Acute Angles

Consider a right triangle. The measure of its right angle is $9 0^{\circ}$ or $\frac{π}{2}$ radians. Let $θ$ be the radian measure of one of the acute angles. Since the sum of two acute angles in a right triangle is $\frac{π}{2},$ the measure of the third acute angle must be $\frac{π}{2} - θ .$

$A right triangle with the side lengths a, b, and c, and two acute angles theta and pi\2-\theta$

Let also

a,

b,

and

c

represent the side lengths of the triangle. In this case, cosine of

θ

can be expressed as the ratio of the lengths of the angle's adjacent side and the hypotenuse.

cos θ = \frac{b}{c}

At the same time, sine of the opposite angle can be expressed as the ratio of the lengths of the angle's opposite side and the hypotenuse.

sin (\frac{π}{2} - θ) = \frac{b}{c}

Since the right-hand sides of the equations are equal, by the Transitive Property of Equality, the left-hand sides are also equal.

$sin (\frac{π}{2} - θ) = cos θ$

This identity is true for all angles, not just those that make it possible to construct a right triangle. Using similar reasoning, the corresponding identities for cosine and tangent can be proven.

Proof

For Any Angle

The given identities can be proven using the Angle Sum and Difference Identities for any angle measure. Consider the identity for the sine of the difference between two angles.

sin (x - y) = sin x cos y - cos x sin y

This identity will be applied to the left-hand side of the first identity.

sin (\frac{π}{2} - θ)

$sin (α - β) = sin (α) cos (β) - cos (α) sin (β)$

sin (\frac{π}{2}) cos (θ) - cos (\frac{π}{2}) sin (θ)

1 \cdot cos (θ) - cos (\frac{π}{2}) sin (θ)

MultByOne

$a \cdot 1 = a$

cos (θ) - cos (\frac{π}{2}) sin (θ)

cos (θ) - 0 \cdot sin (θ)

ZeroPropMult

Zero Property of Multiplication

cos (θ) - 0

IdPropAdd

Identity Property of Addition

cos (θ)

Therefore, the Cofunction Identity for sine was obtained.

$sin (\frac{π}{2} - θ) = cos (θ)$

Rule

Angle Sum and Difference Identities

To evaluate trigonometric functions of the sum of two angles, the following identities can be applied.

sin (x + y) cos (x + y) tan (x + y) = sin x cos y + cos x sin y = cos x cos y - sin x sin y = \frac{tan x + tan y}{1 - tan x tan y}

There are also similar identities for the difference of two angles.

sin (x - y) cos (x - y) tan (x - y) = sin x cos y - cos x sin y = cos x cos y + sin x sin y = \frac{tan x - tan y}{1 + tan x tan y}

These identities are useful when finding the exact value of the sine, cosine, or tangent of a given angle.

Proof

Let $△ A F D$ be a right triangle with hypotenuse $1$ and an acute angle with measure $x + y .$

By definition, the sine of an angle is the ratio between the lengths of the opposite side and the hypotenuse.

sin (x + y) = \frac{D F}{1} ⇓ D F = s i n (x + y)

The idea now is to rewrite

D F

in terms of

sin x,

sin y,

cos x,

and

cos y .

To do it, draw a ray so that

∠ A

is divided into two angles with measures

x

and

y .

Let

C

be a point on this ray such that

△ A C D

and

△ A B C

are right triangles.

Consider

△ A C D .

By calculating the sine and cosine of

x,

the legs of this triangle can be rewritten.

sin x = \frac{D C}{1} \Rightarrow D C = sin x cos x = \frac{A C}{1} \Rightarrow A C = cos x

Now consider

△ A B C .

Knowing that

A C = cos x,

the sine of

y

can be used to write

B C

in terms of

x

and

y .

sin y = \frac{B C}{A C}

▼

Solve for

B C

Substitute

$A C = c o s x$

sin y = \frac{B C}{c o s x}

MultEqn

$LHS \cdot cos x = RHS \cdot cos x$

cos x sin y = B C

RearrangeEqn

Rearrange equation

B C = cos x sin y

Let

G

be the point of intersection between

\overline{F D}

and

\overline{A C} .

Notice that

∠ A G F ≅ ∠ D G C

by the Vertical Angles Theorem.

By the Third Angle Theorem, it is known that $∠ G A F ≅ ∠ G D C .$ Therefore, $m ∠ G D C = y .$

Since the purpose is to rewrite $D F,$ plot a point $E$ on $\overline{D F}$ such that $\overline{E C} ∥ \overline{A B} .$ This way a rectangle $E C B F$ is formed. The opposite sides of a rectangle have the same length, so $E F$ and $C B$ are equal. Also, $\overline{C E} ⊥ \overline{D F}$ makes $△ C E D$ a right triangle.

Consequently,

E F = c o s x s i n y

and

D E

can be written in terms of

sin x

and

cos y

using the cosine ratio.

cos y = \frac{D E}{sin x} ⇓ D E = s i n x c o s y

Finally, by the Segment Addition Postulate,

D F

is equal to the sum of

D E

and

E F .

All these lengths have been rewritten in terms of the sine and cosine of

x

and

y .

D F = D E + E F ⇓ s i n (x + y) = s i n x c o s y + c o s x s i n y

This concludes the proof of the first identity. The other identities can be proven using similar reasoning.

Extra

Calculating

sin 12 0^{\circ}

Consider the following process for calculating the exact value of $sin 12 0^{\circ} .$

To be able to use the angle sum identities, the angle $12 0^{\circ}$ needs to be rewritten as the sum of two angles for which the sine and cosine are known. For example, $12 0^{\circ}$ can be rewritten as $9 0^{\circ} + 3 0^{\circ} .$
Use the first formula for the angle sum.
Based on the trigonometric ratios of common angles, it is known that $sin 9 0^{\circ} = 1,$ $sin 3 0^{\circ} = \frac{1}{2},$ $cos 9 0^{\circ} = 0,$ and $cos 3 0^{\circ} = \frac{3}{2} .$

Following these three steps, the value of

sin 12 0^{\circ}

can be found.

sin 12 0^{\circ}

Rewrite

Rewrite $12 0^{\circ}$ as $9 0^{\circ} + 3 0^{\circ}$

sin (9 0^{\circ} + 3 0^{\circ})

$sin (x + y) = sin x cos y + cos x sin y$

sin 9 0^{\circ} \cdot cos 3 0^{\circ} + cos 9 0^{\circ} \cdot sin 3 0^{\circ}

SubstituteValues

Substitute values

1 \cdot \frac{3}{2} + 0 \cdot \frac{1}{2}

▼

Simplify

OneMult

$1 \cdot a = a$

\frac{3}{2} + 0 \cdot \frac{1}{2}

ZeroPropMult

Zero Property of Multiplication

\frac{3}{2} + 0

IdPropAdd

Identity Property of Addition

\frac{3}{2}

Notice that

12 0^{\circ}

could also be rewritten as

6 0^{\circ} + 6 0^{\circ},

because

sin 6 0^{\circ}

and

cos 6 0^{\circ}

are known values.

Rule

Double-Angle Identities

The double-angle identities materialize when two angles with the same measure are substituted into the angle sum identities.

Sine : Cosine : Tangent : sin 2 θ = 2 sin θ cos θ cos 2 θ = cos^{2} θ - sin^{2} θ cos 2 θ = 2 cos^{2} θ - 1 cos 2 θ = 1 - 2 sin^{2} θ tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}

These identities simplify calculations when evaluating trigonometric functions of twice an angle measure.

Proof

Double-Angle Identities

Start by writing the Angle Sum Identity for sine and cosine.

sin (x + y) = sin x cos y + cos x sin y cos (x + y) = cos x cos y - sin x sin y

Let

x = θ

and

y = θ .

With this,

x + y

becomes

2 θ .

Then, these two formulas can be rewritten in terms of

θ .

sin 2 θ = sin θ cos θ + cos θ sin θ cos 2 θ = cos θ cos θ - sin θ sin θ

Sine Identity

Approaching the first equation, the Commutative Property of Multiplication can be applied to the second term of its right-hand side. Then, by adding the terms on the right-hand side of this equation, the formula for $sin 2 θ$ is obtained.

sin 2 θ = sin θ cos θ + sin θ cos θ ⇓ sin 2 θ = 2 sin θ cos θ ✓

Cosine Identities

Approaching the second equation, the Product of Powers Property can be used to rewrite its right-hand side. By doing this, the first identity for the cosine of the double of an angle is obtained.

cos 2 θ = cos θ cos θ - sin θ sin θ ⇓ cos 2 θ = cos^{2} θ - sin^{2} θ ✓

Now, recall that, by the Pythagorean Identity, the sine square plus the cosine square of the same angle equals

1 .

From this identity, two different equations can be set.

sin^{2} θ + cos^{2} θ = 1 ⇓ {sin^{2} θ = 1 - cos^{2} θ cos^{2} θ = 1 - sin^{2} θ (I) (II)

Next, substitute Equation (I) into the first identity for the cosine.

cos 2 θ = cos^{2} θ - sin^{2} θ

▼

Substitute

1 - cos^{2} θ

for

sin^{2} θ

and simplify

Substitute

$sin^{2} θ = 1 - c o s^{2} θ$

cos 2 θ = cos^{2} θ - (1 - c o s^{2} θ)

Distr

Distribute $- 1$

cos 2 θ = cos^{2} θ - 1 + cos^{2} θ

AddTerms

Add terms

cos 2 θ = 2 cos^{2} θ - 1 ✓

That way, the second identity for the cosine has been obtained. To obtain the third cosine identity, substitute Equation (II) into the first identity for the cosine.

cos 2 θ = cos^{2} θ - sin^{2} θ

▼

Substitute

1 - sin^{2} θ

for

cos^{2} θ

and simplify

Substitute

$cos^{2} θ = 1 - s i n^{2} θ$

cos 2 θ = 1 - s i n^{2} θ - sin^{2} θ

SubTerms

Subtract terms

cos 2 θ = 1 - 2 sin^{2} θ ✓

Tangent Identity

To prove the tangent identity, start by rewriting

tan 2 θ

in terms of sine and cosine.

tan 2 θ = \frac{sin 2 θ}{cos 2 θ}

Next, substitute the first sine identity in the numerator and the first cosine identity in the denominator.

tan 2 θ = \frac{2 sin θ cos θ}{cos ^{2} θ - sin ^{2} θ}

Then, divide the numerator and denominator by

cos^{2} θ .

tan 2 θ = \frac{\frac{2 s i n θ c o s θ}{c o s ^{2} θ}}{\frac{c o s ^{2} θ - s i n ^{2} θ}{c o s ^{2} θ}}

Finally, simplifying the right-hand side the tangent identity will be obtained.

tan 2 θ = \frac{\frac{2 s i n θ c o s θ}{c o s ^{2} θ}}{\frac{c o s ^{2} θ - s i n ^{2} θ}{c o s ^{2} θ}}

▼

Simplify right-hand side

WriteDiffFrac

Write as a difference of fractions

tan 2 θ = \frac{\frac{2 s i n θ c o s θ}{c o s ^{2} θ}}{\frac{c o s ^{2} θ}{c o s ^{2} θ} - \frac{s i n ^{2} θ}{c o s ^{2} θ}}

CrossCommonFac

Cross out common factors

tan 2 θ = \frac{\frac{2 s i n θ c o s θ}{c o s ^{2} θ}}{\frac{c o s ^{2} θ}{c o s ^{2} θ} - \frac{s i n ^{2} θ}{c o s ^{2} θ}}

CancelCommonFac

Cancel out common factors

tan 2 θ = \frac{\frac{2 s i n θ}{c o s θ}}{1 - \frac{s i n ^{2} θ}{c o s ^{2} θ}}

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

tan 2 θ = \frac{\frac{2 s i n θ}{c o s θ}}{1 - ( \frac{s i n θ}{c o s θ} ) ^{2}}

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

tan 2 θ = \frac{2 \cdot \frac{s i n θ}{c o s θ}}{1 - ( \frac{s i n θ}{c o s θ} ) ^{2}}

$\frac{sin ( θ )}{cos ( θ )} = tan (θ)$

tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ} ✓

Extra

Calculating

cos 12 0^{\circ}

To calculate the exact value of $cos 12 0^{\circ},$ these steps can be followed.

To be able to use the double-angle identities, the angle $12 0^{\circ}$ needs to be rewritten as $2$ multiplied by another angle. Therefore, rewrite $12 0^{\circ}$ as $2 \cdot 6 0^{\circ} .$
Use the second formula for the cosine of twice an angle.
Based on the trigonometric ratios of common angles, it is known that $cos 6 0^{\circ} = \frac{1}{2} .$

Following these three steps, the value of

cos 12 0^{\circ}

can be found.

cos 12 0^{\circ}

Rewrite

Rewrite $12 0^{\circ}$ as $2 \cdot 6 0^{\circ}$

cos (2 \cdot 6 0^{\circ})

$cos 2 θ = 2 cos^{2} θ - 1$

2 cos^{2} (6 0^{\circ}) - 1

$cos 6 0^{\circ} = \frac{1}{2}$

2 (\frac{1}{2})^{2} - 1

▼

Simplify

CalcPow

Calculate power

2 \cdot \frac{1}{4} - 1

Multiply

\frac{1}{2} - 1

Rewrite

Rewrite $1$ as $\frac{2}{2}$

\frac{1}{2} - \frac{2}{2}

SubFrac

Subtract fractions

\frac{1 - 2}{2}

SubTerms

Subtract terms

\frac{- 1}{2}

MoveNegNumToFrac

Put minus sign in front of fraction

- \frac{1}{2}

Rule

Half-Angle Identities

The half-angle identities are special cases of angle difference identities. To evaluate trigonometric functions of half an angle, the following identities can be applied.

sin \frac{θ}{2} cos \frac{θ}{2} tan \frac{θ}{2} = \pm \frac{1 - cos θ}{2} = \pm \frac{1 + cos θ}{2} = \pm \frac{1 - cos θ}{1 + cos θ}

The sign of each formula is determined by the quadrant where the angle $\frac{θ}{2}$ lies.

These identities are useful when finding the exact value of the sine, cosine, or tangent at a given angle.

Proof

Half-Angle Identities

First, write two of the Double-Angle Identities for cosine.

cos 2 x cos 2 x = 1 - 2 sin^{2} x = 2 cos^{2} x - 1

Sine Identity

Start by solving the first identity written above for

sin x .

cos 2 x = 1 - 2 sin^{2} x

▼

Solve for

sin x

SubEqn

$LHS - 1 = RHS - 1$

cos 2 x - 1 = - 2 sin^{2} x

DivEqn

$LHS / (- 2) = RHS / (- 2)$

\frac{cos 2 x - 1}{- 2} = sin^{2} x

RearrangeEqn

Rearrange equation

sin^{2} x = \frac{cos 2 x - 1}{- 2}

MoveNegDenomToFrac

Put minus sign in front of fraction

sin^{2} x = - \frac{cos 2 x - 1}{2}

DistrNegSignSwap

$- (b - a) = a - b$

sin^{2} x = \frac{1 - cos 2 x}{2}

SqrtEqn

$LHS = RHS$

sin x = \pm \frac{1 - cos 2 x}{2}

Next, substitute

\frac{θ}{2}

for

x

to obtain the half-angle identity for the sine.

sin \frac{θ}{2} sin \frac{θ}{2} = \pm \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{2} ⇓ = \pm \frac{1 - cos θ}{2} ✓

Cosine Identity

Start by solving the second identity written at the beginning for

cos x .

cos 2 x = 2 cos^{2} x - 1

▼

Solve for

cos x

AddEqn

$LHS + 1 = RHS + 1$

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

DivEqn

$LHS / 2 = RHS / 2$

\frac{1 + cos 2 x}{2} = cos^{2} x

RearrangeEqn

Rearrange equation

cos^{2} x = \frac{1 + cos 2 x}{2}

SqrtEqn

$LHS = RHS$

cos x = \pm \frac{1 + cos 2 x}{2}

Next, substitute

\frac{θ}{2}

for

x

to obtain the half-angle identity for the cosine.

cos \frac{θ}{2} cos \frac{θ}{2} = \pm \frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2} ⇓ = \pm \frac{1 + cos θ}{2} ✓

Tangent Identity

To derive the tangent identity, start by recalling the definition of the tangent ratio.

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

Next, substitute

\frac{θ}{2}

for

x .

tan \frac{θ}{2} = \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

Finally, substitute the half-identities for the sine and cosine into the equation above and simplify the right-hand side.

tan \frac{θ}{2} = \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

SubstituteII

$sin \frac{θ}{2} = \pm \frac{1 - c o s θ}{2}$ , $cos \frac{θ}{2} = \pm \frac{1 + c o s θ}{2}$

tan \frac{θ}{2} = \frac{\pm \frac{1 - c o s θ}{2}}{\pm \frac{1 + c o s θ}{2}}

▼

Simplify right-hand side

DivSqrt

$\frac{a}{b} = \frac{a}{b}$

tan \frac{θ}{2} = \pm \frac{\frac{1 - c o s θ}{2}}{\frac{1 + c o s θ}{2}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

tan \frac{θ}{2} = \pm \frac{1 - cos θ}{2} \cdot \frac{2}{1 + cos θ}

CrossCommonFac

Cross out common factors

tan \frac{θ}{2} = \pm \frac{1 - cos θ}{2} \cdot \frac{2}{1 + cos θ}

CancelCommonFac

Cancel out common factors

tan \frac{θ}{2} = \pm \frac{1 - cos θ}{1} \cdot \frac{1}{1 + cos θ}

MultFrac

Multiply fractions

tan \frac{θ}{2} = \pm \frac{1 - cos θ}{1 + cos θ} ✓

Extra

Calculating

cos 1 5^{\circ}

Consider the calculation of the exact value of $cos 1 5^{\circ} .$

To be able to use the half-angle identities, the angle $1 5^{\circ}$ needs to be rewritten as a certain angle divided by $2 .$ Therefore, rewrite $1 5^{\circ}$ as $\frac{3 0 ^{\circ}}{2} .$
Based on the trigonometric ratios of common angles, it is known that $cos 3 0^{\circ} = \frac{3}{2} .$
According to the diagram of the quadrants, an angle that measures $1 5^{\circ}$ is in the first quadrant. Therefore, the cosine ratio is positive.

With these three steps and the second identity in mind, the value of

cos 1 5^{\circ}

can be found.

cos 1 5^{\circ}

Rewrite

Rewrite $1 5^{\circ}$ as $\frac{3 0 ^{\circ}}{2}$

cos \frac{3 0 ^{\circ}}{2}

$cos \frac{θ}{2} = \frac{1 + cos θ}{2}$

\frac{1 + cos 3 0 ^{\circ}}{2}

$cos 3 0^{\circ} = \frac{3}{2}$

\frac{1 + \frac{3}{2}}{2}

▼

Simplify

OneToFrac

Rewrite $1$ as $\frac{2}{2}$

\frac{\frac{2}{2} + \frac{3}{2}}{2}

AddFrac

Add fractions

\frac{\frac{2 + 3}{2}}{2}

MoveNumRight

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

\frac{1}{2} (\frac{2 + 3}{2})

MultFrac

Multiply fractions

\frac{2 + 3}{4}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

\frac{2 + 3}{4}

CalcRoot

Calculate root

\frac{2 + 3}{2}

As already mentioned, the positive sign was chosen because

1 5^{\circ}

lies in the first quadrant where cosine is positive.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

Proof

Proof

Right Triangle

Unit Circle

Proof

Right Triangle

Unit Circle

Proof

Proof

Proof

Proof

Proof

Proof

Proof

Extra

Proof

Sine Identity

Cosine Identities

Tangent Identity

Extra

Proof

Sine Identity

Cosine Identity

Tangent Identity

Extra