Unlock Trigonometric Identities with Double-Angle and Half-Angle Formulas

$sin^{2} θ = 1 - c o s^{2} θ$

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

$cos^{2} θ = 1 - s i n^{2} θ$

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}}

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}}

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

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}

Subtract fractions

\frac{1 - 2}{2}

SubTerms

Subtract terms

\frac{- 1}{2}

MoveNegNumToFrac

Put minus sign in front of fraction

- \frac{1}{2}

Example

Searching for Clues

When the class got to the first mathematical sacred site, they were told to look for three clues. After eagerly searching the neighborhood, they found three parts to one task.

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The task says to calculate

sin 2 θ

knowing that

cos θ = \frac{2}{5}

and

0^{\circ} \leq θ \leq 9 0^{\circ} .

What exact value should the class get to be able to move on to the next site?

Hint

Find the value of $sin θ$ by using one of the Pythagorean Identities.

Solution

In order to find the values of

sin 2 θ,

recall the Double-Angle Identity for sine.

sin 2 θ = 2 sin θ cos θ

The value of

cos θ

is known, but the value of

sin θ

is not. To find it, one of the Pythagorean Identities can be used.

sin^{2} θ + cos^{2} θ = 1

Substitute

cos θ

with

\frac{2}{5}

and solve for

sin θ .

sin^{2} θ + cos^{2} θ = 1

$cos θ = \frac{2}{5}$

sin^{2} θ + (\frac{2}{5})^{2} = 1

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

sin^{2} θ + \frac{2 ^{2}}{5 ^{2}} = 1

Calculate power

sin^{2} θ + \frac{4}{2 5} = 1

$LHS - \frac{4}{2 5} = RHS - \frac{4}{2 5}$

sin^{2} θ = 1 - \frac{4}{2 5}

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

sin^{2} θ = \frac{2 5}{2 5} - \frac{4}{2 5}

Subtract fractions

sin^{2} θ = \frac{2 1}{2 5}

▼

Calculate root

$LHS = RHS$

∣ sin θ ∣ = \frac{2 1}{2 5}

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

∣ sin θ ∣ = \frac{2 1}{2 5}

Calculate root

∣ sin θ ∣ = \frac{2 1}{5}

By the definition of absolute value,

sin θ

can have two possible values.

∣ sin θ ∣ = \frac{2 1}{5} ⇓ sin θ = \frac{2 1}{5} sin θ = - \frac{2 1}{5}

It is given that

θ

is between

0^{\circ}

and

9 0^{\circ},

which is the first quadrant. Sine has positive values in the first quadrant, so the negative value can be disregarded.

sin θ = \frac{2 1}{5}

Now that both sine and cosine of

θ

are known, the value of

sin 2 θ

can be calculated.

sin 2 θ = 2 sin θ cos θ

$sin θ = \frac{2 1}{5}$ , $cos θ = \frac{2}{5}$

sin 2 θ = 2 (\frac{2 1}{5}) (\frac{2}{5})

Multiply fractions

sin 2 θ = 2 (\frac{2 2 1}{2 5})

MoveLeftFacToNum

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

sin 2 θ = \frac{4 2 1}{2 5}

Therefore, the value of

sin 2 θ

\frac{4 2 1}{2 5} .

Example

Finding Trigonometric Values to Determine the Password

At the second site, a steel safe awaits them. It is locked! Inside is the paper with the information needed to get to the third site. To open the safe, they must figure out the password.

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The password is decoded sequentially by the integers that appear in the answers to the three given tasks. It is known that $sin θ = \frac{1}{3}$ and $9 0^{\circ} \leq θ \leq 18 0^{\circ} .$

a Find the exact value of

sin 2 θ .

b Find the exact value of

cos 2 θ .

c Find the exact value of

tan 2 θ .

Hint

a Use the Pythagorean Identity to find the value of

cos θ .

b Recall the Double-Angle Identity for cosine.

c Apply the Tangent Identity.

Solution

a Start by recalling the Double-Angle Identity for sine.

sin 2 θ = 2 sin θ cos θ

As can be seen, to find the value of

sin 2 θ,

the value of

sin θ

and

cos θ

should be known. It is given that

sin θ = \frac{1}{3} .

Substitute that value into the Pythagorean Identity to calculate the corresponding value of

cos θ .

sin^{2} θ + cos^{2} θ = 1

$sin θ = \frac{1}{3}$

(\frac{1}{3})^{2} + cos^{2} θ = 1

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

\frac{1 ^{2}}{3 ^{2}} + cos^{2} θ = 1

Calculate power

\frac{1}{9} + cos^{2} θ = 1

$LHS - \frac{1}{9} = RHS - \frac{1}{9}$

cos^{2} θ = 1 - \frac{1}{9}

Rewrite $1$ as $\frac{9}{9}$

cos^{2} θ = \frac{9}{9} - \frac{1}{9}

Subtract fractions

cos^{2} θ = \frac{8}{9}

$LHS = RHS$

∣ cos θ ∣ = \frac{8}{9}

▼

Calculate root

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

∣ cos θ ∣ = \frac{8}{9}

SplitIntoFactors

Split into factors

∣ cos θ ∣ = \frac{4 \cdot 2}{9}

Calculate root

∣ cos θ ∣ = \frac{2 2}{3}

By the definition of absolute value, the cosine of

θ

can have two possible values — positive and negative.

∣ cos θ ∣ = \frac{2 2}{3} ⇓ cos θ = \pm \frac{2 2}{3}

It is known that

9 0^{\circ} \leq θ \leq 18 0^{\circ},

which indicates that

θ

is in the second quadrant where cosine has negative values. This way the exact value of

cos θ

is found.

cos θ = - \frac{2 2}{3}

Now, the values of

sin θ

and

cos θ

can be used to calculate the value of

sin 2 θ .

sin 2 θ = 2 sin θ cos θ

$sin θ = \frac{1}{3}$ , $cos θ = - \frac{2 2}{3}$

sin 2 θ = 2 (\frac{1}{3}) (- \frac{2 2}{3})

MultPosNeg

$a (- b) = - a \cdot b$

sin 2 θ = - 2 (\frac{1}{3}) (\frac{2 2}{3})

Multiply fractions

sin 2 θ = - 2 (\frac{2 2}{9})

MoveLeftFacToNum

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

sin 2 θ = - \frac{4 2}{9}

b In order to find the value of

cos 2 θ,

rewrite it by using the Double-Angle Identity for cosine.

cos 2 θ = cos^{2} θ - sin^{2} θ

Since the values of both

sin θ

and

cos θ

are known, substitute them into the equation and solve for

cos 2 θ .

cos 2 θ = cos^{2} θ - sin^{2} θ

$sin θ = \frac{1}{3}$ , $cos θ = - \frac{2 2}{3}$

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

NegBaseToPosBase

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

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

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

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

Calculate power

cos 2 θ = \frac{4 \cdot 2}{9} - \frac{1}{9}

Multiply

cos 2 θ = \frac{8}{9} - \frac{1}{9}

Subtract fractions

cos 2 θ = \frac{7}{9}

c Finally, to find the value of

tan 2 θ,

use the Tangent Identity.

tan 2 θ = \frac{sin 2 θ}{cos 2 θ}

Substitute the found values of

sin 2 θ

and

cos 2 θ

from the previous parts and then solve for

tan 2 θ .

tan 2 θ = \frac{sin 2 θ}{cos 2 θ}

$sin 2 θ = - \frac{4 2}{9}$ , $cos 2 θ = \frac{7}{9}$

tan 2 θ = \frac{- \frac{4 2}{9}}{\frac{7}{9}}

DivFracByFracSL

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

tan 2 θ = - \frac{4 2}{9} \cdot \frac{9}{7}

Multiply fractions

tan 2 θ = - \frac{4 2 \cdot 9}{9 \cdot 7}

Cross out common factors

tan 2 θ = - \frac{4 2 \cdot 9}{9 \cdot 7}

Cancel out common factors

tan 2 θ = - \frac{4 2}{7}

Discussion

Presenting Half-Angle Identities

Before moving to the next station, the class stopped at another infopoint to learn about the Half-Angle Identities.

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

$LHS - 1 = RHS - 1$

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}

$LHS = RHS$

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}

$LHS = RHS$

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} )}

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

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}

Cross out common factors

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

Cancel out common factors

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

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

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})

Multiply fractions

\frac{2 + 3}{4}

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

\frac{2 + 3}{4}

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.

Example

Solving the Tasks Found Inside the Balloons

Zosia and her classmates reached the third site, ready for any task. Three balloons floated steadily in mist. They caught the balloons one by one and noticed that there is something inside each one. Zosia found a needle and popped the balloons.

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Three notes fell to the ground. The classmates have been tasked with finding the values of the following trigonometric expressions by using the Half-Angle Identities. The answers must be written without any radicals in the denominators.

sin 1 5^{\circ}

cos 22 . 5^{\circ}

tan (- 1 5^{\circ})

Hint

a Use the Half-Angle Identity for sine.

b Represent

22 . 5^{\circ}

as a quotient of some angle and

2 .

c Apply the Half-Angle Identity for tangent and the Negative Angle Identity for cosine.

Solution

a First, review the Half-Angle Identity for sine.

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

Note that

15

can be expressed as

\frac{3 0}{2} .

What is more,

θ = 3 0^{\circ}

is a common angle for which trigonometric values are known. Substitute

3 0^{\circ}

for

θ

into the formula and solve for

sin 1 5^{\circ} .

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

$θ = 3 0^{\circ}$

sin (\frac{3 0 ^{\circ}}{2}) = \pm \frac{1 - cos 3 0 ^{\circ}}{2}

CalcQuot

Calculate quotient

sin 1 5^{\circ} = \pm \frac{1 - cos 3 0 ^{\circ}}{2}

sin 1 5^{\circ} = \pm \frac{1 - \frac{3}{2}}{2}

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

sin 1 5^{\circ} = \pm \frac{\frac{2}{2} - \frac{3}{2}}{2}

Subtract fractions

sin 1 5^{\circ} = \pm \frac{\frac{2 - 3}{2}}{2}

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

sin 1 5^{\circ} = \pm \frac{2 - 3}{4}

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

sin 1 5^{\circ} = \pm \frac{2 - 3}{4}

Calculate root

sin 1 5^{\circ} = \pm \frac{2 - 3}{2}

The angle of

1 5^{\circ}

is in the first quadrant where sine has positive values. Therefore,

sin 1 5^{\circ}

has the following value.

sin 1 5^{\circ} = \frac{2 - 3}{2}

b To calculate the cosine of

22 . 5^{\circ},

use the Half-Angle Identity for cosine.

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

The angle of

22 . 5^{\circ}

is equal to

\frac{4 5 ^{\circ}}{2} .

Therefore, substitute

θ

with

4 5^{\circ}

and solve for

cos 22 . 5^{\circ} .

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

$θ = 4 5^{\circ}$

cos (\frac{4 5 ^{\circ}}{2}) = \pm \frac{1 + cos 4 5 ^{\circ}}{2}

CalcQuot

Calculate quotient

cos 22 . 5^{\circ} = \pm \frac{1 + cos 4 5 ^{\circ}}{2}

cos 22 . 5^{\circ} = \pm \frac{1 + \frac{2}{2}}{2}

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

cos 22 . 5^{\circ} = \pm \frac{\frac{2}{2} + \frac{2}{2}}{2}

AddFrac

Add fractions

cos 22 . 5^{\circ} = \pm \frac{\frac{2 + 2}{2}}{2}

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

cos 22 . 5^{\circ} = \pm \frac{2 + 2}{4}

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

cos 22 . 5^{\circ} = \pm \frac{2 + 2}{4}

Calculate root

cos 22 . 5^{\circ} = \pm \frac{2 + 2}{2}

The angle of

22 . 5^{\circ}

is in the first quadrant where the values of cosine are positive. Therefore,

cos 22 . 5^{\circ}

has a positive sign.

cos 22 . 5^{\circ} = \frac{2 + 2}{2}

c The value of

tan (- 1 5^{\circ})

can be calculated by applying the Half-Angle Identity for tangent.

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

Note that

- 1 5^{\circ}

can be represented as

\frac{- 3 0 ^{\circ}}{2} .

Therefore, substitute

θ

with

- 3 0^{\circ} .

tan (\frac{- 3 0}{2}) = \pm \frac{1 - cos ( - 3 0 ^{\circ} )}{1 + cos ( - 3 0 ^{\circ} )}

By the Negative Angle Identity for cosine, the cosine of a negative angle equals the cosine of the positive angle. With this in mind, rewrite the obtained expression.

tan (\frac{- 3 0}{2}) = \pm \frac{1 - cos 3 0 ^{\circ}}{1 + cos 3 0 ^{\circ}}

Simplify the equation and calculate the value of

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

tan (\frac{- 3 0}{2}) = \pm \frac{1 - cos 3 0 ^{\circ}}{1 + cos 3 0 ^{\circ}}

CalcQuot

Calculate quotient

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

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

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

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

AddSubFrac

Add and subtract fractions

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

DivFracByFracSL

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

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

Multiply fractions

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

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

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

Finally, to avoid having a radical in the denominator, the fraction can be multiplied by

(2 - 3) .

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

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 2 - 3 )}{b \cdot ( 2 - 3 )}$

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

▼

Simplify right-hand side

ExpandDiffSquares

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

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

ProdToPowTwoFac

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

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

ExpandNegPerfectSquare

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

tan (- 1 5^{\circ}) = \pm \frac{2 ^{2} - 4 3 + ( 3 ) ^{2}}{2 ^{2} - ( 3 ) ^{2}}

Calculate power

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

AddSubTerms

Add and subtract terms

tan (- 1 5^{\circ}) = \pm \frac{7 - 4 3}{1}

DivByOne

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

tan (- 1 5^{\circ}) = \pm 7 - 43

The angle of

- 1 5^{\circ}

is in the fourth quadrant where tangent is negative. Therefore,

tan (- 1 5^{\circ})

has a negative sign.

tan (- 1 5^{\circ}) = - 7 - 43

Example

Solving a Task from a Fortune Cookie

The class has successfully made it to the fourth mathematical sacred site which appears to be in a kitchen. Something seems strange. They door immediately close behind them. They are not in a kitchen at all, but an escape room! In order to get out, the class needed to find the task and solve it.

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After searching for a while, someone found a fortune cookie. Inside of it, the task challenges them to find the exact value of two expressions given that $sin θ = - \frac{1 5}{1 7}$ and $27 0^{\circ} \leq θ \leq 36 0^{\circ} .$

sin \frac{θ}{2}

cos \frac{θ}{2}

Hint

a Start by calculating

cos θ

by using the Pythagorean Identity.

b Use the Half-Angle Identity for cosine.

Solution

a In order to find the value of

sin \frac{θ}{2},

the Half-Angle Identity for sine can be used.

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

First, the value of the cosine of

θ

needs to be found. To do so, use the Pythagorean Identity.

sin^{2} θ + cos^{2} θ = 1

$sin θ = - \frac{1 5}{1 7}$

(- \frac{1 5}{1 7})^{2} + cos^{2} θ = 1

▼

Solve for

cos θ

NegBaseToPosPow

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

(\frac{1 5}{1 7})^{2} + cos^{2} θ = 1

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

\frac{1 5 ^{2}}{1 7 ^{2}} + cos^{2} θ = 1

Calculate power

\frac{2 2 5}{2 8 9} + cos^{2} θ = 1

$LHS - \frac{2 2 5}{2 8 9} = RHS - \frac{2 2 5}{2 8 9}$

cos^{2} θ = 1 - \frac{2 2 5}{2 8 9}

Rewrite $1$ as $\frac{2 8 9}{2 8 9}$

cos^{2} θ = \frac{2 8 9}{2 8 9} - \frac{2 2 5}{2 8 9}

Subtract fractions

cos^{2} θ = \frac{6 4}{2 8 9}

$LHS = RHS$

∣ cos θ ∣ = \frac{6 4}{2 8 9}

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

∣ cos θ ∣ = \frac{6 4}{2 8 9}

Calculate root

∣ cos θ ∣ = \frac{8}{1 7}

By the definition of absolute value,

cos θ

can have a positive or a negative value.

∣ cos θ ∣ = \frac{8}{1 7} ⇓ cos θ = \pm \frac{8}{1 7}

Recall that

θ

is between

27 0^{\circ}

and

36 0^{\circ},

which is the fourth quadrant where the cosine is positive. This way the exact value of

cos θ

is found.

cos θ = \frac{8}{1 7}

Now, substitute this value into the identity written earlier and evaluate

sin \frac{θ}{2} .

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

$cos θ = \frac{8}{1 7}$

sin \frac{θ}{2} = \pm \frac{1 - \frac{8}{1 7}}{2}

Rewrite $1$ as $\frac{1 7}{1 7}$

sin \frac{θ}{2} = \pm \frac{\frac{1 7}{1 7} - \frac{8}{1 7}}{2}

Subtract fractions

sin \frac{θ}{2} = \pm \frac{\frac{9}{1 7}}{2}

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

sin \frac{θ}{2} = \pm \frac{9}{3 4}

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

sin \frac{θ}{2} = \pm \frac{9}{3 4}

Calculate root

sin \frac{θ}{2} = \pm \frac{3}{3 4}

Angle

θ

is between

27 0^{\circ}

and

36 0^{\circ},

which means that

\frac{θ}{2}

is between

\frac{2 7 0 ^{\circ}}{2} = 13 5^{\circ}

and

\frac{3 6 0 ^{\circ}}{2} = 18 0^{\circ} .

These angles belong to the second quadrant. Therefore, the value of the sine of

\frac{θ}{2}

is positive.

sin \frac{θ}{2} = \frac{3}{3 4}

b To find the value of

cos \frac{θ}{2},

recall the Half-Angle Identity for cosine.

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

From Part A, the value of

cos θ

is known. Substitute it into the formula and solve for

cos \frac{θ}{2} .

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

$cos θ = \frac{8}{1 7}$

cos \frac{θ}{2} = \pm \frac{1 + \frac{8}{1 7}}{2}

Rewrite $1$ as $\frac{1 7}{1 7}$

cos \frac{θ}{2} = \pm \frac{\frac{1 7}{1 7} + \frac{8}{1 7}}{2}

AddFrac

Add fractions

cos \frac{θ}{2} = \pm \frac{\frac{2 5}{1 7}}{2}

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

cos \frac{θ}{2} = \pm \frac{2 5}{3 4}

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

cos \frac{θ}{2} = \pm \frac{2 5}{3 4}

Calculate root

cos \frac{θ}{2} = \pm \frac{5}{3 4}

Earlier it was found that

\frac{θ}{2}

is in the second quadrant where cosine is negative. Therefore, the value of

cos \frac{θ}{2}

is negative.

cos \frac{θ}{2} = - \frac{5}{3 4}

Example

Choosing the Right Direction

The class successfully solved the task from the fortune cookie and arrived at the fifth site. There they saw two potential roads to the next site. To determine which road to choose, they need to solve the task written on the placard.

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Which way should the class turn to end up at the sixth site?

Hint

Use the Pythagorean Identity and the Double-Angle Identity for sine.

Solution

To verify the identity, rewrite its sides until they match. First, add

sin^{2} 2 x

and subtract

4 cos^{4} x

from both sides of the equation. Then, factor out

4 cos^{2} x .

4 cos^{2} x - sin^{2} 2 x = 4 cos^{4} x

AddEqn

$LHS + sin^{2} 2 x = RHS + sin^{2} 2 x$

4 cos^{2} x = 4 cos^{4} x + sin^{2} 2 x

$LHS - 4 cos^{4} x = RHS - 4 cos^{4} x$

4 cos^{2} x - 4 cos^{4} x = sin^{2} 2 x

FactorOut

Factor out $4 cos^{2} x$

4 cos^{2} x (1 - cos^{2} x) = sin^{2} 2 x

Recall the Pythagorean Identity and rewrite it to match the expression in the parentheses.

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

Now, substitute

1 - cos^{2} x

for

sin^{2} x

into the equation and simplify it. Next, use the Double-Angle Identity for sine.

4 cos^{2} x (1 - cos^{2} x) = sin^{2} 2 x

$1 - cos^{2} (θ) = sin^{2} (θ)$

4 cos^{2} x sin^{2} x = sin^{2} 2 x

$sin^{2} (θ) = (sin (θ))^{2}$

4 cos^{2} x sin^{2} x = (sin 2 x)^{2}

$sin (2 θ) = 2 sin (θ) cos (θ)$

4 cos^{2} x sin^{2} x = (2 sin x cos x)^{2}

PowProd

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

4 cos^{2} x sin^{2} x = 4 sin^{2} x cos^{2} x

As can be seen, the left-hand side is the same as the right-hand side. Therefore, the identity is verified. This means that the class should turn right to end up at the sixth site.

Example

Matching the Sides of Identities

The class made the correct turn to the right and arrived at the sixth site. They notice a rustic table with torn pieces of paper, on which the left-hand and right-hand sides of different identities were written. The task at hand is to match at least one of the identities.

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Zosia placed two random pieces in the middle of the table and started checking whether they form an identity.

sin \frac{θ}{2} cos \frac{θ}{2} = ? \frac{sin θ}{2}

What result is she going to get?

Hint

Rewrite one or both sides of the equation by using the Half-Angle Identity or the Double-Angle Identity.

Solution

In order to check whether the equation Zosia formed is an identity, rewrite the sides until they match. There are two ways to do this — using the Half-Angle Identity or using the Double-Angle Identity. Each way will be shown one at a time.

Using the Half-Angle Identity

Start by recalling the Half-Angle Identities for sine and cosine.

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

Substitute the expressions corresponding to

sin \frac{θ}{2}

and

cos \frac{θ}{2}

into the equation and simplify.

sin \frac{θ}{2} cos \frac{θ}{2} = ? \frac{sin θ}{2}

SubstituteExpressions

Substitute expressions

\pm \frac{1 - cos θ}{2} (\pm \frac{1 + cos θ}{2}) = ? \frac{sin θ}{2}

ProdSqrt

$a \cdot b = a \cdot b$

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

Multiply fractions

\pm \frac{( 1 - cos θ ) ( 1 + cos θ )}{4} = ? \frac{sin θ}{2}

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

\pm \frac{1 ^{2} - cos ^{2} θ}{4} = ? \frac{sin θ}{2}

BaseOne

$1^{a} = 1$

\pm \frac{1 - cos ^{2} θ}{4} = ? \frac{sin θ}{2}

Next, the Pythagorean Identity can be used.

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

Substitute

sin^{2} θ

for

1 - cos^{2} θ

into the obtained equation and calculate the square root on the left-hand side of the equation.

\pm \frac{1 - cos ^{2} θ}{4} = ? \frac{sin θ}{2}

$1 - cos^{2} (θ) = sin^{2} (θ)$

\pm \frac{sin ^{2} θ}{4} = ? \frac{sin θ}{2}

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

\pm \frac{sin ^{2} θ}{4} = ? \frac{sin θ}{2}

Calculate root

\pm \frac{sin θ}{2} = \frac{sin θ}{2} ✓

The equation is true, which means that it is an identity.

Using the Double-Angle Identity

Start by reviewing the Double-Angle Identity.

sin 2 θ = 2 sin θ cos θ

If instead of

2 θ

there was

θ,

the identity would look the following way.

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

Finally, by dividing both sides of the identity by

2,

the equation that should have been verified can be obtained.

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

Therefore, the equation Zosia formed is indeed an identity.

Example

Simplifying Trigonometric Expressions

The class has arrived at the seventh mathematical sacred sit — the finale. A huge board with plenty of colorful stickers lies ahead. They are told to choose two random cards.

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The classmates go through and and turn the cards. Lo and behold, two trigonometric expressions appear that need to be simplified. Once this is done, they will have completed the math quest!

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

\frac{cos 2 θ}{sin θ + cos θ}

Hint

a Start by factoring out

- 1 .

b Use the Double-Angle Identity.

Solution

a Start by analyzing the first given expression.

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

As can be seen, the expression contains the squares of sine and cosine. Therefore, it can be simplified by using the Double-Angle Identity for cosine.

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

FactorOut

Factor out $- 1$

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

$cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)$

- cos θ

Therefore, the expression simplifies to

- cos θ .

b Again, to simplify the expression, use the Double-Angle Identity for cosine.

\frac{cos 2 θ}{sin θ + cos θ}

$cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)$

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

FacDiffSquares

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

\frac{( cos θ + sin θ ) ( cos θ - sin θ )}{sin θ + cos θ}

AssociativePropAdd

Associative Property of Addition

\frac{( cos θ + sin θ ) ( cos θ - sin θ )}{cos θ + sin θ}

Cross out common factors

\frac{( cos θ + sin θ ) ( cos θ - sin θ )}{( cos θ + sin θ )}

Cancel out common factors

\frac{cos θ - sin θ}{1}

DivByOne

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

cos θ - sin θ

They did it. They completed the quest!

Closure

Solving the Task to Get the Bonus

When the math quest game for Zosia and her classmates began, they were going to face some tough tasks. After passing all the sites, the rules stated that they could try for a bonus task and get major points. Naturally, they decided try to solve it.

mljsxgraph-error-image-file-not-found:

What is the value of

\frac{H}{D} ?

Hint

Use the Double-Angle Identity for sine and the Tangent Identity.

Solution

To find the value of

\frac{H}{D},

substitute the corresponding expressions to

H

and

D

and then simplify.

\frac{H}{D}

$H = \frac{v ^{2}}{2 g} s i n^{2} θ$ , $D = \frac{v ^{2}}{g} s i n 2 θ$

\frac{\frac{v ^{2}}{2 g} s i n ^{2} θ}{\frac{v ^{2}}{g} s i n 2 θ}

MoveRightFacToNum

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

\frac{\frac{v ^{2} sin ^{2} θ}{2 g}}{\frac{v ^{2} sin 2 θ}{g}}

DivFracByFracD

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

\frac{v ^{2} sin ^{2} θ}{2 g} \cdot \frac{g}{v ^{2} sin 2 θ}

Multiply fractions

\frac{v ^{2} sin ^{2} θ g}{2 g v ^{2} sin 2 θ}

Cross out common factors

\frac{v ^{2} sin ^{2} θ g}{2 g v ^{2} sin 2 θ}

Cancel out common factors

\frac{sin ^{2} θ}{2 sin 2 θ}

Now, the Double-Angle Identity for sine can be used.

\frac{sin ^{2} θ}{2 sin 2 θ}

$sin (2 θ) = 2 sin (θ) cos (θ)$

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

Multiply

\frac{sin ^{2} θ}{4 sin θ cos θ}

Cross out common factors