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$sin^{2} θ = 1 - c o s^{2} θ$

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}}

CrossCommonFac

Cross out common factors

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

CancelCommonFac

Cancel out common factors

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

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

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

MovePartNumLeft

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

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

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

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

Extra

Calculating

cos 12 0^{\circ}

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

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

Following these three steps, the value of

cos 12 0^{\circ}

can be found.

cos 12 0^{\circ}

Rewrite

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

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

$cos 2 θ = 2 cos^{2} θ - 1$

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

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

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

▼

Simplify

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.

Three pieces of paper that say: 'Find sin(2theta)', 'cos(theta)=2/5','theta is between 0 and 90 degrees'

External credits: @brgfx

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

PowQuot

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

sin^{2} θ + \frac{2 ^{2}}{5 ^{2}} = 1

Calculate power

sin^{2} θ + \frac{4}{2 5} = 1

SubEqn

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

sin^{2} θ = 1 - \frac{4}{2 5}

OneToFrac

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}

SqrtQuot

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

∣ sin θ ∣ = \frac{2 1}{2 5}

CalcRoot

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})

MultFrac

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.

External credits: @brgfx, @macrovector

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

PowQuot

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

\frac{1 ^{2}}{3 ^{2}} + cos^{2} θ = 1

Calculate power

\frac{1}{9} + cos^{2} θ = 1

SubEqn

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

cos^{2} θ = 1 - \frac{1}{9}

OneToFrac

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

SqrtQuot

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

∣ cos θ ∣ = \frac{8}{9}

SplitIntoFactors

Split into factors

∣ cos θ ∣ = \frac{4 \cdot 2}{9}

CalcRoot

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})

MultFrac

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}

PowQuot

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

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}

MultFrac

Multiply fractions

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

CrossCommonFac

Cross out common factors

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

CancelCommonFac

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

SubEqn

$LHS - 1 = RHS - 1$

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

DivEqn

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

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

RearrangeEqn

Rearrange equation

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

MoveNegDenomToFrac

Put minus sign in front of fraction

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

DistrNegSignSwap

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

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

$LHS = RHS$

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}