Double angle identities, finally demystified

By MathSolver Editorial Team · Published October 8, 2026 · Updated October 8, 2026

Published October 2026.

Double-angle identities let you collapse 2θ back down to θ, turning an unfamiliar angle into one you already know how to handle. They show up everywhere: simplifying integrals, solving trig equations, and proving other identities. The good news is that all three come from a single idea, so there is really only one thing to memorize.

The one idea behind all three

Every double-angle identity is just an angle-sum identity with both angles equal. You already know sin(A + B) = sin A cos B + cos A sin B. Set B = A, and the right side becomes sin A cos A + cos A sin A = 2 sin A cos A. That is the whole derivation, the other two fall out the same way from cos(A + B) and tan(A + B).

The three identities

Sine: sin 2θ = 2 sin θ cos θ. One form only, clean and memorable.

Cosine, three interchangeable forms:
cos 2θ = cos²θ − sin²θ  =  2cos²θ − 1  =  1 − 2sin²θ.
The first comes straight from cos(A + B); the other two use sin²θ + cos²θ = 1 to eliminate one term. Having all three matters: pick the form whose squared term matches what your problem gives you.

Tangent: tan 2θ = 2tan θ ÷ (1 − tan²θ), valid wherever the denominator is nonzero (θ ≠ 45° + k·90°).

Worked examples

Example 1, exact value. Find sin 2θ given sin θ = 3/5 with θ acute. First get cos θ = 4/5 (3-4-5 triangle). Then sin 2θ = 2 · (3/5) · (4/5) = 24/25. Check it live with our sin cos tan calculator.

Example 2, choosing the cosine form. Simplify 1 − 2sin²(15°). That is exactly the third form of cos 2θ with θ = 15°, so it equals cos 30° = √3/2 ≈ 0.866. Recognizing the pattern backwards is the skill examiners test.

Example 3, solving. Solve sin 2x = sin x for x in [0, 2π). Rewrite: 2 sin x cos x − sin x = 0 → sin x (2 cos x − 1) = 0. So sin x = 0 gives x = 0, π, 2π; cos x = 1/2 gives x = π/3, 5π/3. Five solutions, the double-angle step is what unlocks the factorization.

When to use which cosine form

Given sin θ only → use 1 − 2sin²θ. Given cos θ only → use 2cos²θ − 1. Proving an identity with both → start from cos²θ − sin²θ. This one decision removes most of the flailing students do with these formulas.

Common mistakes

sin 2θ ≠ 2 sin θ. The most damaging error in trigonometry. sin 60° = √3/2 ≈ 0.866, but 2 sin 30° = 1, not equal. The doubling happens inside the function, and the identity tells you what that equals.

cos 2θ ≠ 2 cos θ for the same reason, and cos 2θ ≠ cos²θ either. Write the full identity every time until it is automatic.

Sign errors in tan 2θ, the denominator is 1 minus tan²θ. Flipping it gives the wrong sign for every answer.

Power reduction: running the identities backwards

The cosine forms rearrange into power-reduction identities, the versions calculus actually uses:

sin²θ = (1 − cos 2θ)/2  ·  cos²θ = (1 + cos 2θ)/2

Why they matter: integrating sin²x directly is awkward, but (1 − cos 2x)/2 integrates term by term. Whenever you see a squared trig function inside an integral or a series, power-reduction is the first move. Example: ∫sin²x dx = ∫(1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C.

Deriving cos 2θ’s three forms step by step

Start: cos(A + B) = cos A cos B − sin A sin B. Set B = A: cos 2A = cos²A − sin²A. Form 1 done.

Now use sin²A = 1 − cos²A: cos²A − (1 − cos²A) = 2cos²A − 1. Form 2 done.

Or use cos²A = 1 − sin²A: (1 − sin²A) − sin²A = 1 − 2sin²A. Form 3 done.

Three lines, zero memorization burden, and now you can regenerate any form on demand. Try covering the right-hand sides and re-deriving them; two repetitions usually lock it in permanently.

Cheat sheet

sin 2t = 2 sin t cos t (only form)
cos 2t = cos2t – sin2t = 2cos2t – 1 = 1 – 2sin2t (pick by what is given)
tan 2t = 2tan t / (1 – tan2t)
Power reduction: sin2t = (1 – cos 2t)/2, cos2t = (1 + cos 2t)/2

Practice: test yourself

1. Given cos θ = 5/13 with θ acute, find sin 2θ. (First find sin θ = 12/13, then 2·(12/13)·(5/13) = 120/169.)
2. Simplify cos²(22.5°) − sin²(22.5°). (That is cos 45° = √2/2.)
3. Solve 2cos²x − 1 = 0 for x in [0, π). (Recognize cos 2x = 0 → 2x = π/2, 3π/2 → x = π/4, 3π/4.)

Frequently asked questions

sin 2θ = 2 sin θ cos θ. It comes from the angle-sum formula sin(A+B) with B = A. Example: sin 60° = 2·sin 30°·cos 30° = 2·(1/2)·(√3/2) = √3/2.

They are algebraically equivalent via sin²θ + cos²θ = 1: cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ. Three forms exist so you can pick whichever matches the information your problem gives.

Start from the sum formulas, sin(A+B), cos(A+B), tan(A+B), and set B = A. Simplify, and the double-angle identities drop out. It is a two-line proof once the sum formulas are known.

Simplifying trig expressions, solving equations like sin 2x = cos x, evaluating integrals of sin²x and cos²x (via the power-reduction rearrangements), and proving further identities.