Double Angle Identities

Start by writing the Angle Sum Identity for sine and cosine.

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

Let

x = θ

and

y = θ .

With this,

x + y

becomes

2 θ .

Then, these two formulas can be rewritten in terms of

θ .

sin 2 θ = sin θ cos θ + cos θ sin θ cos 2 θ = cos θ cos θ - sin θ sin θ

Sine Identity

Approaching the first equation, the Commutative Property of Multiplication can be applied to the second term of its right-hand side. Then, by adding the terms on the right-hand side of this equation, the formula for $sin 2 θ$ is obtained.

sin 2 θ = sin θ cos θ + sin θ cos θ ⇓ sin 2 θ = 2 sin θ cos θ ✓

Cosine Identities

Approaching the second equation, the Product of Powers Property can be used to rewrite its right-hand side. By doing this, the first identity for the cosine of the double of an angle is obtained.

cos 2 θ = cos θ cos θ - sin θ sin θ ⇓ cos 2 θ = cos^{2} θ - sin^{2} θ ✓

Now, recall that, by the Pythagorean Identity, the sine square plus the cosine square of the same angle equals

1 .

From this identity, two different equations can be set.

sin^{2} θ + cos^{2} θ = 1 ⇓ {sin^{2} θ = 1 - cos^{2} θ cos^{2} θ = 1 - sin^{2} θ (I) (II)

Next, substitute Equation (I) into the first identity for the cosine.

cos 2 θ = cos^{2} θ - sin^{2} θ

Substitute

1 - cos^{2} θ

for

sin^{2} θ

and simplify

Substitute

$sin^{2} θ = 1 - c o s^{2} θ$

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

Substitute

$cos^{2} θ = 1 - s i n^{2} θ$

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} ✓

{{ article.displayTitle }}

Rule

Double-Angle Identities

Proof

Sine Identity

Cosine Identities

Tangent Identity

Extra

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}