A double angle formula is a trig identity that rewrites a function of 2θ in terms of θ, such as sin(2θ) = 2sinθcosθ and cos(2θ) = cos²θ − sin²θ. In AP Precalculus, you get them by plugging α = β into the sum identities (Topic 3.12) and use them to solve trig equations.
A double angle formula rewrites a trig function of a doubled angle, like sin(2θ), in terms of the original angle θ. The two you need are sin(2θ) = 2sinθcosθ and cos(2θ) = cos²θ − sin²θ. Here's the part that makes them easy to remember instead of memorize: they're not new identities at all. Take the sum identity sin(α + β) = sinα cosβ + cosα sinβ and set α = β = θ. You get sin(θ + θ) = sinθcosθ + cosθsinθ = 2sinθcosθ. Do the same with the cosine sum identity and you get cos(2θ) = cos²θ − sin²θ. The CED says this directly: the sum identities "can also be used as difference and double-angle identities."
The cosine version is the overachiever of the group. Because sin²θ + cos²θ = 1 (the Pythagorean identity), you can swap out either squared term and get two more forms, cos(2θ) = 2cos²θ − 1 and cos(2θ) = 1 − 2sin²θ. Same identity, three outfits. Which form you pick depends on what the problem hands you. If an equation already has sinθ in it, the 1 − 2sin²θ form turns everything into one variable.
Double angle formulas live in Topic 3.12 (Equivalent Representations of Trigonometric Functions) in Unit 3, supporting learning objective AP Pre Calc 3.12.B (rewrite expressions with sine and cosine sum identities) and AP Pre Calc 3.12.C (solve equations using equivalent forms). The whole point of 3.12 is that a trig expression can wear different equivalent forms, and the right form "can make information more accessible." Double angle formulas are the workhorse example. An equation like sin(2x) = cos(x) looks unsolvable because it mixes two different angles. Rewrite sin(2x) as 2sin(x)cos(x) and suddenly everything is in terms of x, you can factor, and the solutions fall out. That rewrite-then-solve move is exactly what 3.12.C asks you to do.
Keep studying AP® Precalculus Unit 3
Sum identity for sine (Unit 3)
The double angle formula IS the sum identity with both angles set equal. sin(2θ) is just sin(θ + θ) expanded. If you know the sum identities, you never have to separately memorize the double angle formulas, you can rebuild them in five seconds.
Pythagorean identity (Unit 3)
The Pythagorean identity sin²θ + cos²θ = 1 (LO 3.12.A) is what generates the alternate forms of cos(2θ). Substituting cos²θ = 1 − sin²θ into cos²θ − sin²θ gives you 1 − 2sin²θ. The two identities work as a team.
Solving trigonometric equations (Unit 3)
Under LO 3.12.C, double angle formulas are an equation-solving tool, not just algebra trivia. They convert a mixed-angle equation into a single-angle one, often something you can factor like 2sinx cosx − cosx = 0, then solve each factor on the unit circle.
Expect multiple-choice questions in two flavors. The first is straight recognition, like "Which formula represents the double angle identity for sine?" or "Which is a valid double angle identity for cosine?" The cosine version is the trap-rich one because three correct forms exist (cos²θ − sin²θ, 2cos²θ − 1, 1 − 2sin²θ) and wrong answers love to flip a sign or swap sine and cosine. The second flavor is application. You'll see an equation like 2sinθcosθ = 1/2 that you should recognize as sin(2θ) = 1/2, or sin(2x) = sinx that you should expand and factor. The skill being tested is choosing the equivalent form that makes the equation solvable, which is the heart of LO 3.12.C.
These aren't rivals, one is a special case of the other. The sum identity sin(α + β) = sinα cosβ + cosα sinβ handles any two angles. The double angle formula is what you get when those two angles happen to be the same. The common mistake runs the other direction, assuming sin(2θ) = 2sin(θ). That's false, and the double angle formula exists precisely because you can't distribute through a trig function. Doubling the angle does not double the output.
The double angle formula for sine is sin(2θ) = 2sinθcosθ, and it comes from setting α = β in the sum identity sin(α + β) = sinα cosβ + cosα sinβ.
The double angle formula for cosine has three equivalent forms, cos(2θ) = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ, thanks to the Pythagorean identity.
sin(2θ) is NOT equal to 2sin(θ); you cannot distribute a trig function over its angle, which is the whole reason these formulas exist.
On the exam, the main job of a double angle formula is converting a mixed-angle equation into a single-angle equation you can factor and solve (LO 3.12.C).
When solving an equation, pick the cos(2θ) form that matches what's already there, like 1 − 2sin²θ if the equation contains sinθ.
It's a trig identity that rewrites a function of 2θ in terms of θ. The sine version is sin(2θ) = 2sinθcosθ and the cosine version is cos(2θ) = cos²θ − sin²θ, which also equals 2cos²θ − 1 and 1 − 2sin²θ. It's part of Topic 3.12 in Unit 3.
No. Trig functions don't distribute over their inputs. The correct expansion is sin(2x) = 2sin(x)cos(x). Quick check at x = 90°: sin(180°) = 0, but 2sin(90°) = 2, so they're clearly not equal.
The double angle formula is just the sum identity with both angles equal. Plug α = β = θ into sin(α + β) = sinα cosβ + cosα sinβ and you get sin(2θ) = 2sinθcosθ. Memorize the sum identities and you get the double angle formulas for free.
Because the Pythagorean identity sin²θ + cos²θ = 1 lets you substitute into cos²θ − sin²θ. Replace sin²θ to get 2cos²θ − 1, or replace cos²θ to get 1 − 2sin²θ. All three are equivalent, and you choose whichever matches the other terms in your equation.
You need to be able to produce them, but the smart route is deriving them from the sum identities, which the CED explicitly says serve as double-angle identities. Knowing the derivation also protects you from sign-flip trap answers on multiple choice.
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