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Trigonometric functions are often used to perform calculations in real-life fields, such as architecture, optics, and trajectories. In many cases, multiple trigonometric functions appear in one expression, potentially causing confusion. Fortunately, rules that relate different trigonometric functions exist, providing the ability to simplify calculations. This lesson will present some of them.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Identity
Unit Circle
Trigonometric Ratios

Challenge

Firework's Height Given by Trigonometric Functions

Best friends Paulina and Maya graduated high school last Friday. At the graduation party, they, along with the rest of the graduates and guests, were admiring bright and colorful fireworks.

External credits: @pch.vector

Their physics teacher came up with a contest to play for a prize — three tickets to an Ali Styles concert! He said that the height of the firework

h

and horizontal displacement

x

were related by the following equation.

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + \frac{x sin θ}{cos θ}

Here,

v

is the initial velocity of the projectile,

θ

is the angle at which it was fired, and

g

is the acceleration due to gravity. The first student, who can rewrite the equation so that

tan θ

is the only trigonometric function, will win the tickets.

Discussion

The Concept of a Trigonometric Identity and Some Examples

Consider the two given equalities. First, analyze if any of them can be simplified. Then, focus on how many values of $x$ make each of them true. What is the main difference between the equalities?

Two equalities: I. x^2+3x=4 and II. x^4/x^3=x

The first equality is an equation that can be solved to find the few, if any, values of $x$ that make the equation true. However, the second equality can be simplified to $x = x$ and, therefore, is an equation that is true for all values of $x$ for which the expressions in the equation are defined. Given that characteristic, the second equality is called an identity.

A trigonometric identity is an equation involving trigonometric functions that is true for all values for which every expression in the equation is defined.

Two of the most basic trigonometric identities are tangent and cotangent Identities. These identities relate tangent and cotangent to sine and cosine.

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.

There are also trigonometric identities which show that some trigonometric functions are reciprocals of others.

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.

Example

Simplifying Expressions by Using Trigonometric Identities

Thinking of different ways to solve the firework challenge, Paulina found herself thinking about her time learning trigonometric identities earlier in the school year. She really enjoyed those lessons.

A few of her favorite exercises included the following where she was asked to simplify these expressions.

\frac{sin ^{2} θ}{tan ^{2} θ}

cot θ sec θ

csc θ tan θ + 3 sec θ

Hint

a Recall the Tangent Identity.

b Use the Cotangent Identity and the Reciprocal Identity for secant.

c Start by simplifying the first term of the sum. Apply the Reciprocal Identity for cosecant and the Tangent Identity.

Solution

a The first expression contains two different trigonometric functions — sine and tangent. This means the Tangent Identity could be useful here.

tan θ = \frac{sin θ}{cos θ}

Substitute

\frac{s i n θ}{c o s θ}

for

tan θ

into the expression and simplify.

\frac{sin ^{2} θ}{tan ^{2} θ}

Substitute

$tan θ = \frac{s i n θ}{c o s θ}$

\frac{sin ^{2} θ}{( \frac{s i n θ}{c o s θ} ) ^{2}}

PowQuot

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

\frac{sin ^{2} θ}{\frac{sin ^{2} θ}{cos ^{2} θ}}

DivByFracD

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

\frac{sin ^{2} θ \cdot cos ^{2} θ}{sin ^{2} θ}

Reduce by

sin^{2} θ

CrossCommonFac

Cross out common factors

\frac{sin ^{2} θ \cdot cos ^{2} θ}{sin ^{2} θ}

CancelCommonFac

Cancel out common factors

\frac{1 \cdot cos ^{2} θ}{1}

OneMult

$1 \cdot a = a$

\frac{cos ^{2} θ}{1}

DivByOne

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

cos^{2} θ

b To simplify the second expression, recall the Cotangent Identity and the Reciprocal Identity for secant.

cot θ = \frac{cos θ}{sin θ} sec θ = \frac{1}{cos θ}

Next, substitute the expressions for

cot θ

and

sec θ

and simplify.

cot θ sec θ

SubstituteII

$cot θ = \frac{c o s θ}{s i n θ}$ , $sec θ = \frac{1}{c o s θ}$

\frac{c o s θ}{s i n θ} \cdot \frac{1}{c o s θ}

MultFrac

Multiply fractions

\frac{cos θ \cdot 1}{sin θ \cdot cos θ}

ReduceFrac

$\frac{a}{b} = \frac{a / cos θ}{b / cos θ}$

\frac{1 \cdot 1}{sin θ \cdot 1}

MultByOne

$a \cdot 1 = a$

\frac{1}{sin θ}

Note that the obtained expression is actually equal to

csc θ .

\frac{1}{sin θ} = csc θ

Therefore, the last expression simplifies to

csc θ .

cot θ sec θ = csc θ

c The last expression is the sum of two terms.

c s c θ t a n θ + 3 s e c θ

Begin by simplifying the

first term .

In order to do this, review the Reciprocal Identity for cosecant and the Tangent Identity.

csc θ = \frac{1}{sin θ} tan θ = \frac{sin θ}{cos θ}

Next, use these identities to find a simpler form of the first term.

csc θ tan θ

SubstituteII

$csc θ = \frac{1}{s i n θ}$ , $tan θ = \frac{s i n θ}{c o s θ}$

\frac{1}{sin θ} \cdot \frac{sin θ}{cos θ}

MultFrac

Multiply fractions

\frac{sin θ}{sin θ cos θ}

CrossCommonFac

Cross out common factors

\frac{sin θ}{sin θ cos θ}

CancelCommonFac

Cancel out common factors

\frac{1}{cos θ}

By the Reciprocal Identity for secant, the obtained expression is equal to

sec θ .

\frac{1}{cos θ} = sec θ

This means that the first term can be simplified to

sec θ .

csc θ tan θ + 3 sec θ

Substitute

$csc θ tan θ = s e c θ$

sec θ + 3 sec θ

AddTerms

Add terms

4 sec θ

Therefore,

4 sec θ

is the most simplified form of the given expression.

Discussion

Suggestions for Verifying Identities

Some techniques, like the following, are helpful when verifying if trigonometric identities are true.

Substitute one or more basic trigonometric identities to simplify the expression.
Factor or multiply as necessary. Sometimes it is necessary to multiply both the numerator and denominator by the same trigonometric expression.
Write each side of the identity in terms of sine and cosine only. Then simplify each side as much as possible.

Verifying an identity tan(theta)=sec(theta)/csc(theta)

Remember that the properties of equality do not apply to identities as they do with equations. It is not possible to perform operations to the quantities on each side of an unverified identity.

Discussion

Presenting Pythagorean Identities

One of the most known trigonometric identities relates the square of sine and cosine of any angle $θ .$ This identity can be manipulated to obtain two more identities involving other trigonometric functions.

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.

Now it will be shown how the Pythagorean Identities can be applied to solve different mathematical problems.

Example

Trigonometric Functions of Angles on the Map

Enjoying the graduation party, Maya and Paulina shared some memories about some of the fun they had exploring their neighborhood as kids. After school they would buy junk food at the

24 -

hour store, bird watch in the trees of the park, and play around the lake.

The map of Paulina's and Maya's neighborhood

External credits: Hari Panicker

The sine of the angle

θ

formed by the road connecting Paulina's and Maya's houses to the store is

\frac{4}{5} .

The cosine of the angle

β

formed by the road connecting Paulina's and Maya's houses to the school is

- \frac{7}{8} .

Interact with the map to view these angles.

a What is the cosine of

θ ?

b What is the tangent of

θ ?

c What is the sine of

β ?

d What is the cotangent of

β ?

Hint

a Apply the Pythagorean Identity.

b Recall the Tangent Identity.

c To determine the sign of the sine, identify in which quadrant

β

is located.

d Use the Cotangent Identity.

Solution

a It is given that the sine of

θ

\frac{4}{5} .

To find the value of the cosine of

θ,

the Pythagorean Identity can be used.

sin^{2} θ + cos^{2} θ = 1

Substitute the known value of

sin θ

and solve for

cos θ .

sin^{2} θ + cos^{2} θ = 1

Substitute

$sin θ = \frac{4}{5}$

(\frac{4}{5})^{2} + cos^{2} θ = 1

Solve for

cos θ

PowQuot

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

\frac{4 ^{2}}{5 ^{2}} + cos^{2} θ = 1

CalcPow

Calculate power

\frac{1 6}{2 5} + cos^{2} θ = 1

SubEqn

$LHS - \frac{1 6}{2 5} = RHS - \frac{1 6}{2 5}$

cos^{2} θ = 1 - \frac{1 6}{2 5}

OneToFrac

Rewrite $1$ as $\frac{2 5}{2 5}$

cos^{2} θ = \frac{2 5}{2 5} - \frac{1 6}{2 5}

SubFrac

Subtract fractions

cos^{2} θ = \frac{9}{2 5}

CalcRoot

Calculate root

cos θ = \pm \frac{3}{5}

To determine the sign of

cos θ,

identify in which quadrant the angle is situated.

External credits: Hari Panicker

The angle is in the first quadrant, where cosine is positive. Therefore, it can be concluded that the cosine of $θ$ is $\frac{3}{5} .$

b To calculate the tangent of

θ,

the Tangent Identity can be used.

tan θ = \frac{sin θ}{cos θ}

Since both

sin θ

and

cos θ

are known, substitute their values and solve for

tan θ .

tan θ = \frac{sin θ}{cos θ}

SubstituteII

$sin θ = \frac{4}{5}$ , $cos θ = \frac{3}{5}$

tan θ = \frac{\frac{4}{5}}{\frac{3}{5}}

Simplify right-hand side

DivFracByFracSL

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

tan θ = \frac{4}{5} \cdot \frac{5}{3}

MultFrac

Multiply fractions

tan θ = \frac{4 \cdot 5}{5 \cdot 3}

CrossCommonFac

Cross out common factors

tan θ = \frac{4 \cdot 5}{5 \cdot 3}

CancelCommonFac

Cancel out common factors

tan θ = \frac{4}{3}

c This time the sine of the angle is known, while the cosine should be found. Again, the Pythagorean Theorem can be used. Substitute

- \frac{7}{8}

for

cos β

and solve for

sin β .

sin^{2} β + cos^{2} β = 1

Substitute

$cos β = - \frac{7}{8}$

sin^{2} β + (- \frac{7}{8})^{2} = 1

Solve for

sin β

NegBaseToPosPow

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

sin^{2} β + (\frac{7}{8})^{2} = 1

PowQuot

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

sin^{2} β + \frac{7 ^{2}}{8 ^{2}} = 1

CalcPow

Calculate power

sin^{2} β + \frac{4 9}{6 4} = 1

SubEqn

$LHS - \frac{4 9}{6 4} = RHS - \frac{4 9}{6 4}$

sin^{2} β = 1 - \frac{4 9}{6 4}

OneToFrac

Rewrite $1$ as $\frac{6 4}{6 4}$

sin^{2} β = \frac{6 4}{6 4} - \frac{4 9}{6 4}

SubFrac

Subtract fractions

sin^{2} β = \frac{1 5}{6 4}

CalcRoot

Calculate root

sin β = \pm \frac{1 5}{8}

Next, identify the quadrant of the angle to determine the sign of the sine of

β .

External credits: Hari Panicker

The angle is in the second quadrant, where sine is positive. Therefore, the sine of $β$ is $\frac{1 5}{8} .$

d In order to find the cotangent of

β,

recall the Cotangent Identity.

cot β = \frac{cos β}{sin β}

Substitute

cos β

with

- \frac{7}{8}

and

sin β

with

\frac{1 5}{8}

and solve for

cot β .

cot β = \frac{cos β}{sin β}

SubstituteII

$cos β = - \frac{7}{8}$ , $sin β = \frac{1 5}{8}$

cot β = \frac{- \frac{7}{8}}{\frac{1 5}{8}}

Simplify right-hand side

DivFracByFracSL

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

cot β = - \frac{7}{8} \cdot \frac{8}{1 5}

MultFrac

Multiply fractions

cot β = - \frac{7 \cdot 8}{8 \cdot 1 5}

CrossCommonFac

Cross out common factors

cot β = - \frac{7 \cdot 8}{8 \cdot 1 5}

CancelCommonFac

Cancel out common factors

cot β = - \frac{7}{1 5}

Pop Quiz

Practicing Applying Pythagorean Identities

Given the value of a trigonometric function and the quadrant of the angle, use one of the Pythagorean Identities to find the value of another trigonometric function. Round the answer to two decimal places.

Applet that generates questions about the value of a trigonometric function

Example

Verifying an Identity

Maya told Jordan, who was late for the graduation party, all about the contest and asked her to recall the trigonometric identities they learned on her way over. While thinking about the identities, Jordan remembered that, when they studied the topic, Maya sent her one interesting identity and decided to find that text.

The phone with text messages from the girls to each other. Maya sent an identity sec(theta)csc(theta)=tan(theta)+cot(theta)

Is the equation Maya sent a trigonometric identity?

Hint

Use the known trigonometric identities to rewrite the left- and right-hand sides of the equation until they match.

Solution

To verify the given identity, its sides should be rewritten until they match. First, rewrite the left-hand side using the Reciprocal Identities for secant and cosecant.

sec θ csc θ = tan θ + cot θ

SubstituteII

$sec θ = \frac{1}{c o s θ}$ , $csc θ = \frac{1}{s i n θ}$

\frac{1}{c o s θ} \cdot \frac{1}{s i n θ} = tan θ + cot θ

MultFrac

Multiply fractions

\frac{1}{cos θ sin θ} = tan θ + cot θ

Next, the right-hand side can be rewritten by applying the Tangent and Cotangent Identities.

\frac{1}{cos θ sin θ} = tan θ + cot θ

SubstituteII

$tan θ = \frac{s i n θ}{c o s θ}$ , $cot θ = \frac{c o s θ}{s i n θ}$

\frac{1}{cos θ sin θ} = \frac{s i n θ}{c o s θ} + \frac{c o s θ}{s i n θ}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot sin θ}{b \cdot sin θ}$

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

ExpandFrac

$\frac{a}{b} = \frac{a \cdot cos θ}{b \cdot cos θ}$

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

AddFrac

Add fractions

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

Finally, by the Pythagorean Identity, the numerator of the fraction on the right-hand side is

1 .

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

A true statement was obtained, which means that the equation Maya sent is, indeed, a trigonometric identity.

Discussion

Cofunction and Negative Angles Identities

Studying angles and their trigonometric values in a right triangle more closely reveals further relationships between sine and cosine, and between tangent and cotangent.

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.

When dealing with negative angles, identities that relate trigonometric values of negative and positive angles become very useful.

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)

Example

Solving an Online Quiz

Jordan, still on the bus ride to the graduation party, feels excited about getting to solve a trig problem to win tickets to see Ali Styles live. She still has a bit of time before arriving so she used her tablet to look up a $10 -$ minute online quiz as practice.

Maya's laptop with the open website with the quiz

Here are the expressions that Jordan is asked to simplify. What answers should be submitted to score a perfect $100 % ?$

cos 5 0^{\circ} cos 4 0^{\circ} - sin 5 0^{\circ} sin 4 0^{\circ}

\frac{sin ( - θ )}{cos ( - θ )}

\frac{cos ( \frac{π}{2} - θ )}{csc θ} + cos^{2} θ

cot (- θ) cot (\frac{π}{2} - θ)

Hint

a Use the Cofunction Identities for sine and cosine.

b Recall the Negative Angle Identities.

c Rewrite the expression by applying the Cofunction Identity for cosine and the Reciprocal Identity for cosecant.

d Begin by using the Cotangent Identity.

Solution

a The first expression consists of two similar terms. To rewrite it, the Cofunction Identities can be used.

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

Recall that

\frac{π}{2}

radians equals

9 0^{\circ} .

Apply each identity to the terms and then simplify.

cos 5 0^{\circ} cos 4 0^{\circ} - sin 5 0^{\circ} sin 4 0^{\circ}

$cos (θ) = sin (9 0^{\circ} - θ)$

sin (9 0^{\circ} - 5 0^{\circ}) cos 4 0^{\circ} - sin 5 0^{\circ} sin 4 0^{\circ}

$sin (θ) = cos (9 0^{\circ} - θ)$

sin (9 0^{\circ} - 5 0^{\circ}) cos 4 0^{\circ} - cos (9 0^{\circ} - 5 0^{\circ}) sin 4 0^{\circ}

SubTerm

Subtract term

sin 4 0^{\circ} cos 4 0^{\circ} - cos 4 0^{\circ} sin 4 0^{\circ}

DiffZero

$A - A = 0$

0

Therefore, the expression simplifies to

0 .

b To simplify the second expression, first recall the Negative Angle Identities.

sin (- θ) cos (- θ) = - sin θ = cos θ

Use them to rewrite the numerator and denominator of the fraction. Then apply the Tangent Identity.

\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}

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

- tan θ

c First, analyze the given expression.

\frac{cos ( \frac{π}{2} - θ )}{csc θ} + cos^{2} θ

In order to simplify it, rewrite the fraction by using the Cofunction Identity for cosine and the Reciprocal Identity for cosecant.

\frac{cos ( \frac{π}{2} - θ )}{csc θ} + cos^{2} θ

$cos (\frac{π}{2} - θ) = sin (θ)$

\frac{sin θ}{csc θ} + cos^{2} θ

$csc (θ) = \frac{1}{sin ( θ )}$

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

DivByFracD

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

\frac{sin θ sin θ}{1} + cos^{2} θ

ProdToPowTwoFac

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

\frac{sin ^{2} θ}{1} + cos^{2} θ

DivByOne

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

sin^{2} θ + cos^{2} θ

By the Pythagorean Identity, the obtained expression equals to

1 .

sin^{2} θ + cos^{2} θ = 1

d Start by rewriting the cotangent by applying the Cotangent Identity.

cot (- θ) cot (\frac{π}{2} - θ)

$cot (θ) = \frac{1}{tan ( θ )}$

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

MultFrac

Multiply fractions

\frac{1}{tan ( - θ ) tan ( \frac{π}{2} - θ )}

Next, use the Cofunction and Negative Angle Identities for tangent.

\frac{1}{tan ( - θ ) tan ( \frac{π}{2} - θ )}

$tan (- θ) = - tan (θ)$

\frac{1}{- tan θ tan ( \frac{π}{2} - θ )}

Substitute

$tan (\frac{π}{2} - θ) = c o t θ$

\frac{1}{- tan θ cot θ}

WriteProdFrac

Write as a product of fractions

\frac{1}{- tan θ} \cdot \frac{1}{cot θ}

MoveNegDenomToFrac

Put minus sign in front of fraction

- \frac{1}{tan θ} \cdot \frac{1}{cot θ}

Finally, the first fraction can be rewritten by applying the Reciprocal Identity for cotangent.

- \frac{1}{tan θ} \cdot \frac{1}{cot θ}

$cot (θ) = \frac{1}{tan ( θ )}$

- cot θ \cdot \frac{1}{cot θ}

DenomMultFracToNumber

$cot θ \cdot \frac{a}{cot θ} = a$

- 1

Therefore, the expression simplifies to

- 1 .

Example

Building a Birdhouse

Back to remembering their fun times before graduation, Paulina and Maya laughed about a walk in the park they had. One day, they noticed a beautiful birdhouse hanging on a maple tree. Multiple birds were eating from it and others were singing! They just had to try and build one themselves.

Birdhouse is hanging on the branch of a tree

External credits: @brgfx

They measured the dimensions of the birdhouse and the angle of its roof incline

θ,

as the applet can display. However, they decided to change one part of the birdhouse and agreed to calculate its length at home. Later, when they compared notes, they saw their results were different.

Paulina ’ s Result : Maya ’ s Result : \frac{3 cos ( 9 0 ^{\circ} - θ )}{1 - cos ( - θ )} 3 (csc θ + cot θ)

Are the girl's results equivalent or did someone make a mistake?

Hint

Start by applying the Cofunction and Negative Angle Identities for cosine. Then find by which factor the fraction should be expanded so that the Pythagorean Identity could be used.

Solution

In order to determine whether the expressions are equivalent, try to rewrite Paulina's expression until it matches Maya's expression.

\frac{3 cos ( 9 0 ^{\circ} - θ )}{1 - cos ( - θ )}

Since the arguments of the cosines are

9 0^{\circ} - θ

and

- θ,

use the Cofunction and Negative Angle Identities for cosine to change the arguments to

θ .

\frac{3 cos ( 9 0 ^{\circ} - θ )}{1 - cos ( - θ )}

$cos (9 0^{\circ} - θ) = sin (θ)$

\frac{3 sin θ}{1 - cos ( - θ )}

$cos (- θ) = cos (θ)$

\frac{3 sin θ}{1 - cos θ}

Next, to be able to apply another trigonometric identity to the denominator, expand the fraction by the factor of

(1 + cos θ) .

\frac{3 sin θ}{1 - cos θ}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 1 + cos θ )}{b \cdot ( 1 + cos θ )}$

\frac{3 sin θ ( 1 + c o s θ )}{( 1 - cos θ ) ( 1 + c o s θ )}

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

\frac{3 sin θ ( 1 + cos θ )}{1 ^{2} - cos ^{2} θ}

BaseOne

$1^{a} = 1$

\frac{3 sin θ ( 1 + cos θ )}{1 - cos ^{2} θ}

Note that the denominator has the term

cos^{2} θ .

This term is also in the Pythagorean Identity. Rewrite this identity to get a similar expression on one side of the identity as the one in the denominator.

sin^{2} θ + cos^{2} θ = 1 ⇓ sin^{2} θ = 1 - cos^{2} θ

The rewritten identity can now be used to simplify the denominator and reduce the fraction by the factor of

sin θ .

Next, distribute

3

in the numerator and split the fraction as a sum of fractions.

\frac{3 sin θ ( 1 + cos θ )}{1 - cos ^{2} θ}

$1 - cos^{2} (θ) = sin^{2} (θ)$

\frac{3 sin θ ( 1 + cos θ )}{sin ^{2} θ}

ReduceFrac

$\frac{a}{b} = \frac{a / sin θ}{b / sin θ}$

\frac{3 ( 1 + cos θ )}{sin θ}

Distr

Distribute $3$

\frac{3 + 3 cos θ}{sin θ}

WriteSumFrac

Write as a sum of fractions

\frac{3}{sin θ} + \frac{3 cos θ}{sin θ}

Finally, notice that the fractions can be rewritten by using the Cotangent Identity and the Reciprocal Identity for cosecant.

\frac{3}{sin θ} + \frac{3 cos θ}{sin θ}

MoveNumLeft

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

3 \frac{1}{sin θ} + \frac{3 cos θ}{sin θ}

$csc (θ) = \frac{1}{sin ( θ )}$

3 csc θ + \frac{3 cos θ}{sin θ}

MovePartNumLeft

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

3 csc θ + 3 \frac{cos θ}{sin θ}

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

3 csc θ + 3 cot θ

FactorOut

Factor out $3$

3 (csc θ + cot θ)

This expression is the same as Maya's result, which means that Paulina's and Maya's expressions are equivalent. Therefore, both of their calculations are correct — they just obtained different forms of the same answer. Eventually, they were able to build an awesome birdhouse. What a memory.

Closure

Trying to Win the Contest

Earlier, it was said that at the graduation party, Paulina and Maya, along with the rest of the graduates and guests, were admiring the bright sparkles of a firework.

External credits: @pch.vector

Their physics teacher came up with a contest to win a prize — three tickets to an Ali Styles concert! The teacher said that the height of the firework

h

and horizontal displacement

x

were related by the following equation.

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + \frac{x sin θ}{cos θ}

Here,

v

is the initial velocity of the projectile,

θ

is the angle at which it was fired, and

g

is the acceleration due to gravity. How can this equation be rewritten so that

tan θ

is the only trigonometric function that appears in the equation?

Hint

Use two of the Reciprocal Identities for tangent and secant. Then apply one of the Pythagorean Identities.

Solution

Paulina and Maya really want to win the prize, so much so that they even sent practice problems to Jordan on her bus ride to the party. Now that Jordan has finally joined, they started analyzing the given equation.

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + \frac{x sin θ}{cos θ}

First, Maya noticed that one of the Reciprocal Identities can be used to rewrite the second fraction in terms of

tan θ .

tan θ = \frac{sin θ}{cos θ}

They decided to substitute it into the equation.

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + \frac{x sin θ}{cos θ}

MovePartNumLeft

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

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + x \frac{sin θ}{cos θ}

Substitute

$\frac{sin θ}{cos θ} = t a n θ$

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + x t a n θ

Remember, the purpose is to rewrite the expression in terms of tangent. Paulina notices that cosine still remains in the equation. It hits her that there is another Reciprocal Identity that relates cosine and secant. Since the equation contains a square of cosine, the girls decided to square each side of the identity.

sec θ = \frac{1}{cos θ} ⇓ sec^{2} θ = \frac{1 ^{2}}{cos ^{2} θ}

Then, they substituted the expression into the equation.

h = \frac{- g x ^{2}}{2 v ^{2} cos ^{2} θ} + x tan θ

WriteProdFrac

Write as a product of fractions

h = \frac{- g x ^{2}}{2 v ^{2}} \cdot \frac{1}{cos ^{2} θ} + x tan θ

Substitute

$\frac{1}{cos ^{2} θ} = s e c^{2} θ$

h = \frac{- g x ^{2}}{2 v ^{2}} (s e c^{2} θ) + x tan θ

Finally, Jordan suggests the idea to use one of the Pythagorean Identities containing secant.

h = \frac{- g x ^{2}}{2 v ^{2}} (sec^{2} θ) + x tan θ

Substitute

$sec^{2} θ = 1 + t a n^{2} θ$

h = \frac{- g x ^{2}}{2 v ^{2}} (1 + t a n^{2} θ) + x tan θ

This way they arrived at the equation where the tangent is the only trigonometric function. Their physics teacher checked their work. The results are in, and they won the tickets to the concert!

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Catch-Up and Review

Proof

Right Triangle

Unit Circle

Proof

Right Triangle

Unit Circle

Proof

Hint

Solution

Proof

Proof

Hint

Solution

Hint

Solution

Proof

Proof

Proof

Hint

Solution

Hint

Solution

Hint

Solution