Double Angle Identities

Learn the double angle formulas for sine, cosine, and tangent to simplify expressions and solve equations.

Advanced30 minLesson
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Definition

The double angle identities express trigonometric functions of 2θ2\theta in terms of functions of θ\theta.

Sine Double Angle

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta

Cosine Double Angle (Three Forms)

cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta
cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1
cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta) = 1 - 2\sin^2\theta

Tangent Double Angle

tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}
These identities are derived from the sum formulas by setting both angles equal to θ\theta.

Try it now

What is the double angle formula for sine?

Worked Examples

Use the sum formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B to derive sin⁡(2θ)\sin(2\theta).
1

Start with the sum formula

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B → Sum formula
2

Set A=B=θA = B = \theta

sin⁡(θ+θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ\sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta → sin⁡(2θ)=sin⁡θcos⁡θ+sin⁡θcos⁡θ\sin(2\theta) = \sin\theta\cos\theta + \sin\theta\cos\theta
3

Combine like terms

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta

Common Mistakes

Writing sin⁡(2θ)=2sin⁡θ\sin(2\theta) = 2\sin\theta

Why it's wrong: Students sometimes forget the cosine factor. The sine function is not linear, so you cannot simply double the argument by doubling the output.

Correct: sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta: both sine AND cosine of θ\theta are needed.

Using the wrong form of cos⁡(2θ)\cos(2\theta)

Why it's wrong: There are three equivalent forms. Choosing the wrong one can make problems harder.

Correct: Choose the form that matches what you know: use 1−2sin⁡2θ1 - 2\sin^2\theta if you know sin⁡θ\sin\theta, use 2cos⁡2θ−12\cos^2\theta - 1 if you know cos⁡θ\cos\theta.

Forgetting the sign of tan⁡(2θ)\tan(2\theta) denominator

Why it's wrong: The formula has 1−tan⁡2θ1 - \tan^2\theta in the denominator, not 1+tan⁡2θ1 + \tan^2\theta.

Correct: tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

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Double Angle, Power Reduction, and Half Angle Identities Made Clear

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Interactive Visual

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sin(θ)

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Click and drag to rotate the angle around the unit circle.

Function graph

-300-200-1000100200300-2-1012xy(30, 0.87)

Tap or hover over the graph to read off values.

  • sin(2 · a) = 0.87
  • 2 · sin(a) · cos(a) = 0.87
  • 2 · sin(a) = 1
y = sin(2x)y = 2sin(x)

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Practice Problems

15 problems
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What is the double angle formula for sine?

Why It Matters

Double angle identities are essential tools in advanced mathematics and physics:
  • Calculus: Simplify integrals like ∫sin⁡2x dx\int \sin^2 x \, dx using the identity sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1 - \cos(2x)}{2}
  • Physics: Describe projectile motion, where range depends on sin⁡(2θ)\sin(2\theta)
  • Signal Processing: Analyze wave interference and modulation
  • Engineering: Design rotating machinery and oscillating systems
Without these identities, many calculations in science and engineering would be far more complex!

Real World Applications

Projectile Range Formula

The range of a projectile launched at angle $\theta$ with initial velocity $v_0$ is $R = \frac{v_0^2 \sin(2\theta)}{g}$. Maximum range occurs when $\sin(2\theta) = 1$, i.e., $\theta = 45°$.

Example:

A football kicked at 20 m/s at a 30° angle: R=400⋅sin⁡(60°)10=400⋅0.86610≈34.6R = \frac{400 \cdot \sin(60°)}{10} = \frac{400 \cdot 0.866}{10} \approx 34.6 meters.

1Try It Yourself

An athlete throws a javelin at 25 m/s. They want to maximize the distance.

At what angle should they throw, and what is the maximum range?

Step 1: Write the mathematical expression

For maximum range, sin⁡(2θ)=1\sin(2\theta) = 1, so 2θ=90°2\theta = 90°

Electrical Engineering: Power Factor

In AC circuits, power calculations often involve $\cos(2\omega t)$ where $\omega$ is angular frequency. The double angle identity helps analyze instantaneous power.

Example:

If voltage V=V0cos⁡(ωt)V = V_0\cos(\omega t) and current I=I0cos⁡(ωt)I = I_0\cos(\omega t), instantaneous power involves cos⁡2(ωt)=1+cos⁡(2ωt)2\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}.

2Try It Yourself

An engineer needs to find the average of cos⁡2(ωt)\cos^2(\omega t) over one period.

Use the identity cos⁡2θ=1+cos⁡(2θ)2\cos^2\theta = \frac{1 + \cos(2\theta)}{2} to find this average.

Step 1: Write the mathematical expression

The average of cos⁡(2ωt)\cos(2\omega t) over a period is 0, so...

Key Takeaways

  • 1sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta
  • 2cos⁡(2θ)\cos(2\theta) has three forms: cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta, 2cos⁡2θ−12\cos^2\theta - 1, and 1−2sin⁡2θ1 - 2\sin^2\theta
  • 3tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}
  • 4These identities come from the sum formulas with A=B=θA = B = \theta
  • 5Choose the form of cos⁡(2θ)\cos(2\theta) that matches your given information

Frequently Asked Questions

All three are equivalent, but each is useful in different situations. Use cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta when you know both. Use 2cos⁡2θ−12\cos^2\theta - 1 when you only know cosine. Use 1−2sin⁡2θ1 - 2\sin^2\theta when you only know sine. They are derived using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
All three are equivalent, but each is useful in different situations. Use cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta when you know both. Use 2cos⁡2θ−12\cos^2\theta - 1 when you only know cosine. Use 1−2sin⁡2θ1 - 2\sin^2\theta when you only know sine. They are derived using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
Replace θ\theta with θ2\frac{\theta}{2} in the double angle formula. For example, from cos⁡(2α)=2cos⁡2α−1\cos(2\alpha) = 2\cos^2\alpha - 1, let α=θ2\alpha = \frac{\theta}{2}: cos⁡θ=2cos⁡2(θ2)−1\cos\theta = 2\cos^2(\frac{\theta}{2}) - 1, which gives cos⁡(θ2)=±1+cos⁡θ2\cos(\frac{\theta}{2}) = \pm\sqrt{\frac{1 + \cos\theta}{2}}.
When the denominator equals zero: 1−tan⁡2θ=01 - \tan^2\theta = 0, so tan⁡θ=±1\tan\theta = \pm 1, meaning θ=45°,135°,225°,315°\theta = 45°, 135°, 225°, 315° (or π4,3π4,5π4,7π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}). At these angles, 2θ2\theta is an odd multiple of 90°90°.

Glossary

Double angle
An angle that is twice another angle, written as 2θ2\theta
Identity
An equation that is true for all valid values of the variable
Sum formula
Formulas like sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A \cos B + \cos A \sin B used to find trig values of sums
Pythagorean identity
The fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Formula Card

Sine

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2\sin\theta\cos\theta

Double angle formula for sine

Cosine (Form 1)

cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta) = \cos^2\theta - \sin^2\theta

Using both sine and cosine

Cosine (Form 2)

cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1

Using only cosine

Cosine (Form 3)

cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta) = 1 - 2\sin^2\theta

Using only sine

Tangent

tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

Double angle formula for tangent

Today's challengeOne problem, three tries

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