Unlocking Trigonometric Identities: Angle Sum and Difference, Notable Angles, and Trigonometric Values

There are identities that relate the trigonometric values of two angles to the trigonometric values of the sum or difference of these two angles. In this lesson, they will be introduced and practiced.

Catch-Up and Review

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

Challenge

Designing a City Park

Tiffaniqua, who works as a landscape designer, received a job to create a new design for an old city park. Since the park is quite huge, she divided its area into six rectangular sections. The first section contains a fountain $(F)$ and is crossed by a river at two points — south $(S)$ and north $(N) .$

The first section of the park with a river and a fountain

When she first came to analyze the park, she stood at the north-west corner of the first section, which she marked as point $M .$ Then, she took notes of some measures of angles and distances.

A piece of paper with the following notes: angle MSF=30 degrees, angle MSN=45 degrees and MS=40 meters

Later when returning to her work space, Tiffaniqua used her notes to make additional calculations. What is the length of the river within the first section of the park? Round the answer to the first decimal place.

Discussion

Introducing Angle Sum and Difference Identities

There are identities that allow calculating the values of trigonometric functions of the sum or difference of two angles.

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}

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.

They can be used to calculate the exact value of the sine, cosine, or tangent of a given angle.

Example

Evaluating Trigonometric Expressions at Random Values

When Tiffaniqua came home from work, she saw that her son Davontay and his friend Zain came up with a game. Davontay assigned numbers $1$ through $6$ to the trigonometric functions of sine, cosine, and tangent, while Zain assigned numbers $1$ through $6$ to six angle measures.

Assigned numbers for the first die: 1-sin(x), 2-cos(x), 3-tan(x), 4-tan(x), 5-sin(x), 6-cos(x); assigned numbers for the second die: 1-75 degrees, 2-(-15)degrees, 3-195 degrees, 4-240 degrees, 5-105 degrees, 6-(-165)degrees

Next, they rolled the dice four times to identify which two trigonometric values each person should calculate. The die on the left determines the trigonometric function and the die on the right determines the angle measure.

Rolling two dice four times results in getting the following four expressions: sin(105), tan(195), cos(75), tan(-15)

Find the exact values of the expressions that Davontay and Zain obtained. Write the answers in such a way that there are no radicals in the denominators.

sin 10 5^{\circ}

tan 19 5^{\circ}

cos 7 5^{\circ}

tan (- 1 5^{\circ})

Hint

a Express

10 5^{\circ}

as the sum or difference of two notable angles.

b Use the Angle Sum and Difference Identities for tangent.

c Note that

7 5^{\circ}

can be expressed as the difference of

12 0^{\circ}

and

4 5^{\circ} .

d Try to simplify the fraction by multiplying both the numerator and denominator by such an expression that would remove a radical from the denominator.

Solution

a The exact value of

sin 10 5^{\circ}

should be found. Recall that the values of the trigonometric functions of some notable angles are known. Try to represent the angle of

10 5^{\circ}

as the sum or difference of two notable angles.

10 5^{\circ} = 6 0^{\circ} + 4 5^{\circ}

Now, use the Angle Sum Identity for sine.

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

Substitute

6 0^{\circ}

for

x

and

4 5^{\circ}

for

y

and simplify.

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

SubstituteII

$x = 6 0^{\circ}$ , $y = 4 5^{\circ}$

sin (6 0^{\circ} + 4 5^{\circ}) = sin 6 0^{\circ} cos 4 5^{\circ} + cos 6 0^{\circ} sin 4 5^{\circ}

AddTerms

Add terms

sin 10 5^{\circ} = sin 6 0^{\circ} cos 4 5^{\circ} + cos 6 0^{\circ} sin 4 5^{\circ}

▼

Substitute values

sin 10 5^{\circ} = \frac{3}{2} cos 4 5^{\circ} + cos 6 0^{\circ} sin 4 5^{\circ}

sin 10 5^{\circ} = \frac{3}{2} \cdot \frac{2}{2} + cos 6 0^{\circ} sin 4 5^{\circ}

sin 10 5^{\circ} = \frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} sin 4 5^{\circ}

sin 10 5^{\circ} = \frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2}

MultFrac

Multiply fractions

sin 10 5^{\circ} = \frac{6}{4} + \frac{2}{4}

AddFrac

Add fractions

sin 10 5^{\circ} = \frac{6 + 2}{4}

b Similarly to Part A, start by expressing

19 5^{\circ}

as the sum or difference of two notable angles.

19 5^{\circ} = 24 0^{\circ} - 4 5^{\circ}

Next, apply the Angle Difference Identity for tangent.

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

Substitute

24 0^{\circ}

for

x

and

4 5^{\circ}

for

y,

and then find the value of

tan 195

by solving the equation.

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

SubstituteII

$x = 24 0^{\circ}$ , $y = 4 5^{\circ}$

tan (24 0^{\circ} - 4 5^{\circ}) = \frac{tan 2 4 0 ^{\circ} - tan 4 5 ^{\circ}}{1 + tan 2 4 0 ^{\circ} tan 4 5 ^{\circ}}

SubTerm

Subtract term

tan 19 5^{\circ} = \frac{tan 2 4 0 ^{\circ} - tan 4 5 ^{\circ}}{1 + tan 2 4 0 ^{\circ} tan 4 5 ^{\circ}}

Substitute

$tan (24 0^{\circ}) = 3$

tan 19 5^{\circ} = \frac{3 - tan 4 5 ^{\circ}}{1 + 3 tan 4 5 ^{\circ}}

tan 19 5^{\circ} = \frac{3 - 1}{1 + 3 \cdot 1}

MultByOne

$a \cdot 1 = a$

tan 19 5^{\circ} = \frac{3 - 1}{1 + 3}

Finally, to simplify the obtained expression, multiply both the numerator and denominator by

(1 - 3) .

tan 19 5^{\circ} = \frac{3 - 1}{1 + 3}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 1 - 3 )}{b \cdot ( 1 - 3 )}$

tan 19 5^{\circ} = \frac{( 3 - 1 ) ( 1 - 3 )}{( 1 + 3 ) ( 1 - 3 )}

▼

Simplify right-hand side

ExpandDiffSquares

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

tan 19 5^{\circ} = \frac{( 3 - 1 ) ( 1 - 3 )}{1 ^{2} - ( 3 ) ^{2}}

MultPar

Multiply parentheses

tan 19 5^{\circ} = \frac{3 - ( 3 ) ^{2} - 1 + 3}{1 ^{2} - ( 3 ) ^{2}}

CalcPow

Calculate power

tan 19 5^{\circ} = \frac{3 - 3 - 1 + 3}{1 - 3}

AddSubTerms

Add and subtract terms

tan 19 5^{\circ} = \frac{2 3 - 4}{- 2}

MoveNegDenomToFrac

Put minus sign in front of fraction

tan 19 5^{\circ} = - \frac{2 3 - 4}{2}

SimpQuot

Simplify quotient

tan 19 5^{\circ} = - (3 - 2)

Distr

Distribute $- 1$

tan 19 5^{\circ} = - 3 + 2

CommutativePropAdd

Commutative Property of Addition

tan 19 5^{\circ} = 2 - 3

c First, note that the angle of

7 5^{\circ}

can be represented as the difference of

12 0^{\circ}

and

4 5^{\circ} .

Therefore, the Angle Difference Identity for cosine can be used to calculate the value of

cos 7 5^{\circ} .

7 5^{\circ} = 12 0^{\circ} - 4 5^{\circ}

Next, use the known values of the trigonometric functions of notable angles

12 0^{\circ}

and

4 5^{\circ} .

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

SubstituteII

$x = 12 0^{\circ}$ , $y = 4 5^{\circ}$

cos (12 0^{\circ} - 4 5^{\circ}) = cos 12 0^{\circ} cos 4 5^{\circ} + sin 12 0^{\circ} sin 4 5^{\circ}

SubTerm

Subtract term

cos (7 5^{\circ}) = cos 12 0^{\circ} cos 4 5^{\circ} + sin 12 0^{\circ} sin 4 5^{\circ}

▼

Substitute values

cos 7 5^{\circ} = - \frac{1}{2} cos 4 5^{\circ} + sin 12 0^{\circ} sin 4 5^{\circ}

cos 7 5^{\circ} = - \frac{1}{2} \cdot \frac{2}{2} + sin 12 0^{\circ} sin 4 5^{\circ}

cos 7 5^{\circ} = - \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} sin 4 5^{\circ}

cos 7 5^{\circ} = - \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} \cdot \frac{2}{2}

MultFrac

Multiply fractions

cos 7 5^{\circ} = - \frac{2}{4} + \frac{6}{4}

CommutativePropAdd

Commutative Property of Addition

cos 7 5^{\circ} = \frac{6}{4} - \frac{2}{4}

SubFrac

Subtract fractions

cos 7 5^{\circ} = \frac{6 - 2}{4}

d To find the value of

tan (- 1 5^{\circ}),

first express the angle of

- 1 5^{\circ}

as the sum or difference of two notable angles.

- 1 5^{\circ} = 3 0^{\circ} - 4 5^{\circ}

Next, use the Angle Difference Identity for tangent. Substitute

x

with

3 0^{\circ}

and

y

with

4 5^{\circ}

and solve the equation for

tan (- 1 5^{\circ}) .

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

SubstituteII

$x = 3 0^{\circ}$ , $y = 4 5^{\circ}$

tan (3 0^{\circ} - 4 5^{\circ}) = \frac{tan 3 0 ^{\circ} - tan 4 5 ^{\circ}}{1 + tan 3 0 ^{\circ} tan 4 5 ^{\circ}}

SubTerm

Subtract term

tan (- 1 5^{\circ}) = \frac{tan 3 0 ^{\circ} - tan 4 5 ^{\circ}}{1 + tan 3 0 ^{\circ} tan 4 5 ^{\circ}}

▼

Simplify right-hand side

tan (- 1 5^{\circ}) = \frac{\frac{3}{3} - tan 4 5 ^{\circ}}{1 + \frac{3}{3} tan 4 5 ^{\circ}}

tan (- 1 5^{\circ}) = \frac{\frac{3}{3} - 1}{1 + \frac{3}{3} \cdot 1}

MultByOne

$a \cdot 1 = a$

tan (- 1 5^{\circ}) = \frac{\frac{3}{3} - 1}{1 + \frac{3}{3}}

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

tan (- 1 5^{\circ}) = \frac{\frac{3}{3} - \frac{3}{3}}{\frac{3}{3} + \frac{3}{3}}

AddFrac

Add fractions

tan (- 1 5^{\circ}) = \frac{\frac{3 - 3}{3}}{\frac{3 + 3}{3}}

DivFracByFracSL

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

tan (- 1 5^{\circ}) = \frac{3 - 3}{3} \cdot \frac{3}{3 + 3}

MultFrac

Multiply fractions

tan (- 1 5^{\circ}) = \frac{3 ( 3 - 3 )}{3 ( 3 + 3 )}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

tan (- 1 5^{\circ}) = \frac{3 - 3}{3 + 3}

The expression can be simplified by multiplying both the numerator and denominator of the fraction by

(3 - 3) .

tan (- 1 5^{\circ}) = \frac{3 - 3}{3 + 3}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 3 - 3 )}{b \cdot ( 3 - 3 )}$

tan (- 1 5^{\circ}) = \frac{( 3 - 3 ) ( 3 - 3 )}{( 3 + 3 ) ( 3 - 3 )}

▼

Simplify right-hand side

ExpandDiffSquares

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

tan (- 1 5^{\circ}) = \frac{( 3 - 3 ) ( 3 - 3 )}{( 3 ) ^{2} - 3 ^{2}}

ProdToPowTwoFac

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

tan (- 1 5^{\circ}) = \frac{( 3 - 3 ) ^{2}}{( 3 ) ^{2} - 3 ^{2}}

ExpandNegPerfectSquare

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

tan (- 1 5^{\circ}) = \frac{( 3 ) ^{2} - 6 3 + 3 ^{2}}{( 3 ) ^{2} - 3 ^{2}}

CalcPow

Calculate power

tan (- 1 5^{\circ}) = \frac{3 - 6 3 + 9}{3 - 9}

AddSubTerms

Add and subtract terms

tan (- 1 5^{\circ}) = \frac{1 2 - 6 3}{- 6}

MoveNegDenomToFrac

Put minus sign in front of fraction

tan (- 1 5^{\circ}) = - \frac{1 2 - 6 3}{6}

SimpQuot

Simplify quotient

tan (- 1 5^{\circ}) = - (2 - 3)

Distr

Distribute $- 1$

tan (- 1 5^{\circ}) = - 2 + 3

CommutativePropAdd

Commutative Property of Addition

tan (- 1 5^{\circ}) = 3 - 2

Example

Evaluating Trigonometric Functions

In the game that Davontay and Zain created and played, Davontay solved everything correctly. Zain, on the other hand, made one mistake. This was on Zain's mind as they came home, so they decided to practice by evaluating more trigonometric functions.

1. sin(23pi/12) 2. cos(-165) 3. tan(255)

Find the values of the given expressions along with Zain.

sin \frac{2 3 π}{1 2}

cos (- 16 5^{\circ})

tan 25 5^{\circ}

Hint

a The angle

\frac{2 3 π}{1 2}

can be expressed as the sum of

\frac{2 0 π}{1 2}

and

\frac{3 π}{1 2} .

b Start by using the Negative Angle Identity for cosine.

c Apply the Angle Sum or Difference Identity for tangent.

Solution

a To calculate the value of

sin \frac{2 3 π}{1 2},

start by expressing the angle of

\frac{2 3 π}{1 2}

as the sum or difference of two notable angles.

\frac{2 3 π}{1 2} = \frac{2 0 π}{1 2} + \frac{3 π}{1 2}

Note that the angles

\frac{2 0 π}{1 2}

and

\frac{3 π}{1 2}

can be simplified to

\frac{5 π}{3}

and

\frac{π}{4},

which are the notable angles corresponding to

30 0^{\circ}

and

4 5^{\circ} .

Next, use the Angle Sum Identity for sine.

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Catch-Up and Review

Proof

Extra

Hint

Solution

Hint

Solution