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

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

Derive an identity for

cot (A - B)

in terms of

cot A

and

cot B .

We are asked to derive an identity for cot(A-B) in terms of cotA and cotB. Let's begin by recalling that the cotangent can be written as the ratio of cosine to sine. cotθ=cosθ/sinθ Since we are looking for an identity for the difference of two angles, we will also review the Angle Difference Identities for sine, cosine, and tangent.

Difference Identities
sin(A-B) = sinAcosB-cosAsinB
cos(A-B) = cosAcosB+sinAsinB
tan(A-B) = tanA-tanB/1-tanAtanB

Let's work the identity!

We will now divide both the numerator and denominator by sinA and sinB. By doing so, we can get cot A and cot B in the numerator and the denominator. Let's start by dividing by sinA.

Likewise, we will divide by sinB in the numerator and denominator and then simplify the expression.

We obtained an identity for cotangent of the difference of two angles. cot(A-B)=cotAcotB+1/cotB-cotA

Consider two angles $a$ and $b$ in standard position and the corresponding points $A$ and $B$ on the unit circle.

Angles a and b along with points A and B on a unit circle

Use the Distance Formula to find

d

in terms of

cos (a - b),

where

(x_{1}, y_{1})

are the coordinates of

B

and

(x_{2}, y_{2})

are the coordinates of

A .

We are given two angles in standard position on the unit circle. We also know that the points A and B lie on the unit circle and that their coordinates are (x_2,y_2) and (x_1,y_1), respectively.

Since the points A and B lie on the unit circle, we can express their coordinates as (cos a,sin a) and (cos b,sin b), respectively.

To find the distance d, we can use the Distance Formula. Let's recall how the formula looks! d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) In our case, (x_1,y_1) are the coordinates of B and (x_2,y_2) are the coordinates of A. (x_1,y_1)=(cos b,sin b) (x_2,y_2)=(cos a,sin a) Let's substitute them into the Distance Formula and simplify. For simplicity, we will raise both sides of the equation to the second power to eliminate the square root.

Now, recall one of the Pythagorean Identities. cos^2 θ+sin^2 θ=1 We can use this identity to simplify our expression.

Notice that inside the parentheses we have the formula for the cosine of the difference of angles a and b. Let's use this fact to further simplify the expression.

The distance d between the points is sqrt(2-2cos(a-b)).

Sum and Difference Angle Identities

Catch-Up and Review

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Introducing Angle Sum and Difference Identities

Angle Sum and Difference Identities

Proof

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Sum and Difference Angle Identities

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