Double-Angle Formulas: sin 2θ, cos 2θ, tan 2θ

The double-angle formulas rewrite trig functions of $2\theta$ in terms of $\theta$ : $\sin 2\theta = 2 \sin \theta \cos \theta$ , $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$ . Cosine also has two other forms, $2\cos^2 \theta - 1$ and $1 - 2\sin^2 \theta$ . These formulas all come from the sum formulas. They help you find exact values, simplify expressions and solve trig equations.

Before this lesson: You should know Sum and Difference Formulas and Fundamental Trig Identities.

Where the Formulas Come From

Doubling an angle does not double its sine. For example, $\sin 60^\circ \approx 0.866$ , but $2 \sin 30^\circ = 1$ . To get the right formula, write $2\theta$ as $\theta + \theta$ and use the sum formula:

\sin(\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta = 2 \sin \theta \cos \theta

The same trick with the cosine sum formula gives $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ . Replacing $\sin^2 \theta$ or $\cos^2 \theta$ with a Pythagorean identity gives the other two forms.

Graph with x from 0 to 2 pi and y from −2.5 to 2.5. A blue solid curve shows y = sin 2x, which repeats twice and stays between −1 and 1. An orange dashed curve shows y = 2 sin x, which repeats once and reaches 2 and −2. The curves cross at 0, pi and 2 pi but are clearly different. — sin 2x is not the same as 2 sin x. Doubling the angle doubles the speed of the wave, not its height.

The Double-Angle Formulas

Function	Formula
$\sin 2\theta$	$2 \sin \theta \cos \theta$
$\cos 2\theta$	$\cos^2 \theta - \sin^2 \theta$
$\cos 2\theta$	$2\cos^2 \theta - 1$
$\cos 2\theta$	$1 - 2\sin^2 \theta$
$\tan 2\theta$	$\frac{2 \tan \theta}{1 - \tan^2 \theta}$

Find $\sin \theta$ , $\cos \theta$ or $\tan \theta$ , using the quadrant to choose signs.
Choose the formula that uses the values you have. For cosine, pick the form with only sine or only cosine if that is all you know.
Substitute and simplify.
Check the sign: the quadrant of $2\theta$ decides the sign of the answer.

Worked Examples

Example 1: All three double-angle values

If $\sin \theta = \frac{3}{5}$ and $\theta$ is in Quadrant I, find $\sin 2\theta$ , $\cos 2\theta$ and $\tan 2\theta$ .

\cos \theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}

Find cosine with the Pythagorean identity.

\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}

Use the sine formula.

\cos 2\theta = (\frac{4}{5})^2 - (\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}

Use the cosine formula.

\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}

Divide for tangent.

Answer: $\sin 2\theta = \frac{24}{25}$ , $\cos 2\theta = \frac{7}{25}$ , $\tan 2\theta = \frac{24}{7}$

Example 2: An angle in Quadrant II

If $\cos \theta = -\frac{5}{13}$ and $\theta$ is in Quadrant II, find $\sin 2\theta$ and $\cos 2\theta$ .

\sin \theta = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}

Find sine, positive in Quadrant II.

\sin 2\theta = 2 \cdot \frac{12}{13} \cdot (-\frac{5}{13})

Use the sine formula.

\sin 2\theta = -\frac{120}{169}

Multiply.

\cos 2\theta = 2(-\frac{5}{13})^2 - 1

Use the cosine-only form.

\cos 2\theta = 2 \cdot \frac{25}{169} - 1 = \frac{50}{169} - \frac{169}{169}

Square.

\cos 2\theta = -\frac{119}{169}

Subtract.

Answer: $\sin 2\theta = -\frac{120}{169}$ and $\cos 2\theta = -\frac{119}{169}$ . Both are negative, so $2\theta$ is in Quadrant III.

Example 3: Collapse an expression

Find the exact values of $2 \sin 15^\circ \cos 15^\circ$ and $\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8}$ .

2 \sin 15^\circ \cos 15^\circ = \sin 30^\circ

The first matches $2 \sin \theta \cos \theta$ with $\theta = 15^\circ$ .

\sin 30^\circ = \frac{1}{2}

Evaluate.

\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8} = \cos \frac{\pi}{4}

The second matches $\cos^2 \theta - \sin^2 \theta$ with $\theta = \frac{\pi}{8}$ .

\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}

Evaluate.

Answer: $\frac{1}{2}$ and $\frac{\sqrt{2}}{2}$

Example 4: Verify an identity

Verify $(\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta$ .

\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta

Start with the left side and expand.

(\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cos \theta

Group the squares.

1 + 2 \sin \theta \cos \theta

Use the Pythagorean identity.

1 + \sin 2\theta

Use the double-angle formula.

Answer: verified, using the method from verifying trig identities.

Example 5: Word problem with a projectile

A ball is kicked at 20 m/s at an angle $\theta$ above the ground. Ignoring air resistance, it lands $R = \frac{v^2 \sin 2\theta}{g}$ meters away, where $v$ is the speed and $g = 9.8$ . Find the distance when $\theta = 15^\circ$ and when $\theta = 45^\circ$ .

\sin 2\theta = \sin 30^\circ = 0.5

For 15°, double the angle.

R = \frac{20^2 \cdot 0.5}{9.8} = \frac{400 \cdot 0.5}{9.8} = \frac{200}{9.8} \approx 20.41

Substitute.

\sin 2\theta = \sin 90^\circ = 1

For 45°, double the angle.

R = \frac{400 \cdot 1}{9.8} \approx 40.82

Substitute.

Answer: about 20.41 m at 15° and 40.82 m at 45°. Because $\sin 2\theta$ is largest when $2\theta = 90^\circ$ , a 45° angle gives the longest kick.

Common Mistakes

Mistake	Example of the error	Fix
Doubling the sine	$\sin 2\theta = 2 \sin \theta$	Use $\sin 2\theta = 2 \sin \theta \cos \theta$ .
Mixing up the cosine forms	$\cos 2\theta = 1 - 2\cos^2 \theta$	The forms are $2\cos^2 \theta - 1$ and $1 - 2\sin^2 \theta$ .
Wrong sign for a missing value	Using $\sin \theta = -\frac{12}{13}$ in Quadrant II	Sine is positive in Quadrant II.
Tangent formula with a plus	$\tan 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$	The denominator is $1 - \tan^2 \theta$ .
Halving the angle when collapsing	$2 \sin 15^\circ \cos 15^\circ = \sin 7.5^\circ$	The formula doubles the angle: $\sin 30^\circ$ .

Related Methods

Power-reducing formulas

Solving $\cos 2\theta = 2\cos^2 \theta - 1$ for $\cos^2 \theta$ gives $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ . Similarly, $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ . These lead to the half-angle formulas in the next lesson.

Triple angles

You can keep going: $\sin 3\theta = \sin(2\theta + \theta)$ . Using the sum formula and the double-angle formulas gives $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$ .

Solving equations

Equations like $\sin 2x = \cos x$ become easier after rewriting $\sin 2x$ as $2 \sin x \cos x$ and factoring. You will do this in the trig equations unit.

Practice Problems

If $\sin \theta = \frac{4}{5}$ and $\theta$ is in Quadrant I, find $\sin 2\theta$ .
Use the angle in problem 1 to find $\cos 2\theta$ .
If $\tan \theta = \frac{1}{2}$ , find $\tan 2\theta$ .
If $\cos \theta = \frac{12}{13}$ and $\theta$ is in Quadrant IV, find $\sin 2\theta$ .
Find the exact value of $2 \sin 22.5^\circ \cos 22.5^\circ$ .
Find the exact value of $1 - 2\sin^2 \frac{\pi}{12}$ .
Find the exact value of $2\cos^2 75^\circ - 1$ .
Simplify $\frac{\sin 2\theta}{2 \sin \theta}$ .
Verify $\cos^4 \theta - \sin^4 \theta = \cos 2\theta$ .
A ball is thrown at 30 m/s at a 30° angle. Use $R = \frac{v^2 \sin 2\theta}{9.8}$ to find how far it travels, to the nearest hundredth.

Show answers

\cos \theta = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

Find cosine.

\sin 2\theta = 2 \cdot \frac{4}{5} \cdot \frac{3}{5}

Substitute.

\frac{24}{25}

Multiply.

Answer: $\frac{24}{25}$

\cos 2\theta = (\frac{3}{5})^2 - (\frac{4}{5})^2

Substitute.

\frac{9}{25} - \frac{16}{25}

Square.

-\frac{7}{25}

Subtract.

Answer: $-\frac{7}{25}$

\tan 2\theta = \frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2}

Substitute.

\frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}}

Simplify the top and bottom.

\frac{4}{3}

Divide.

Answer: $\frac{4}{3}$

\sin \theta = -\sqrt{1 - \frac{144}{169}} = -\frac{5}{13}

Find sine, negative in Quadrant IV.

\sin 2\theta = 2 \cdot (-\frac{5}{13}) \cdot \frac{12}{13}

Substitute.

-\frac{120}{169}

Multiply.

Answer: $-\frac{120}{169}$

\sin(2 \cdot 22.5^\circ) = \sin 45^\circ

It matches $2 \sin \theta \cos \theta$ .

\frac{\sqrt{2}}{2}

Evaluate.

Answer: $\frac{\sqrt{2}}{2}$

\cos(2 \cdot \frac{\pi}{12}) = \cos \frac{\pi}{6}

It matches $1 - 2\sin^2 \theta$ .

\frac{\sqrt{3}}{2}

Evaluate.

Answer: $\frac{\sqrt{3}}{2}$

\cos(2 \cdot 75^\circ) = \cos 150^\circ

It matches $2\cos^2 \theta - 1$ .

-\frac{\sqrt{3}}{2}

Use the reference angle 30° in Quadrant II.

Answer: $-\frac{\sqrt{3}}{2}$

\frac{2 \sin \theta \cos \theta}{2 \sin \theta}

Use the formula.

\cos \theta

Cancel $2 \sin \theta$ .

Answer: $\cos \theta$

(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)

Factor as a difference of squares.

(\cos^2 \theta - \sin^2 \theta) \cdot 1

Use the Pythagorean identity.

\cos 2\theta

Use the double-angle formula.

Answer: verified

10.

\sin 60^\circ \approx 0.86603

Double the angle.

R = \frac{900 \cdot 0.86603}{9.8}

Substitute.

R \approx 79.53

Multiply and divide.

Answer: about 79.53 meters

FAQ

What is the double-angle formula for sine?

$\sin 2\theta = 2 \sin \theta \cos \theta$ . It comes from the sum formula with both angles equal to $\theta$ .

What are the three forms of cos 2θ?

$\cos^2 \theta - \sin^2 \theta$ , $2\cos^2 \theta - 1$ and $1 - 2\sin^2 \theta$ . They are equal because $\sin^2 \theta + \cos^2 \theta = 1$ .

Is sin 2x the same as 2 sin x?

No. For $x = 90^\circ$ , $\sin 180^\circ = 0$ but $2 \sin 90^\circ = 2$ . Doubling the angle is different from doubling the value.

Which cos 2θ formula should I use?

Use the form that matches what you know. If you only know $\cos \theta$ , use $2\cos^2 \theta - 1$ ; if you only know $\sin \theta$ , use $1 - 2\sin^2 \theta$ .

Summary

$\sin 2\theta = 2 \sin \theta \cos \theta$ .
$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta$ .
$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$ .
All of them come from the sum formulas with $A = B = \theta$ .
$\sin 2\theta$ is not $2 \sin \theta$ .

Previous lesson: Sum and Difference Formulas

Next lesson: Half-Angle Formulas

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