Double-angle identities express sin(2θ) and cos(2θ) in terms of trig functions of θ alone, such as sin(2θ) = 2 sin θ cos θ and cos(2θ) = cos²θ − sin²θ. In AP Precalculus, you get them by plugging α = β = θ into the sum identities (Topic 3.12).
Double-angle identities are equivalent forms for the sine and cosine of twice an angle. The big three to know are sin(2θ) = 2 sin θ cos θ, and cos(2θ) = cos²θ − sin²θ, which can be rewritten as 2cos²θ − 1 or 1 − 2sin²θ using the Pythagorean identity.
Here's the part the CED actually cares about. These are not new facts to memorize from scratch. They're the sum identities with both angles set equal. Take sin(α + β) = sin α cos β + cos α sin β, let α = β = θ, and you get sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ. Same move with the cosine sum identity gives cos(2θ) = cos²θ − sin²θ. That's why the CED says the sum identities "can also be used as difference and double-angle identities." One identity, dressed up three ways.
Double-angle identities live in Topic 3.12 (Equivalent Representations of Trigonometric Functions) in Unit 3, and they directly support LO 3.12.B, rewriting trig expressions using the sine and cosine sum identities. They also feed LO 3.12.C, because the whole point of an equivalent form is to make an equation solvable. An equation like cos(2θ) = sin θ looks impossible until you swap cos(2θ) for 1 − 2sin²θ. Suddenly it's a quadratic in sin θ, and you can factor it. The exam tests whether you recognize that a rewrite is the unlock, not just whether you've memorized formulas. The three faces of cos(2θ) also connect back to LO 3.12.A, since converting between them is a one-step application of sin²θ + cos²θ = 1.
Keep studying AP® Precalculus Unit 3
Sum identity for sine (Unit 3)
The double-angle identities ARE the sum identities with α = β. If you know sin(α + β) = sin α cos β + cos α sin β, you can rebuild sin(2θ) = 2 sin θ cos θ in one line, so you never have to trust your memory under exam pressure.
Pythagorean identity (Unit 3)
sin²θ + cos²θ = 1 is what converts cos(2θ) = cos²θ − sin²θ into its other two forms, 2cos²θ − 1 and 1 − 2sin²θ. Multiple-choice questions love asking which form is NOT equivalent, and the Pythagorean identity is how you check.
Solving trigonometric equations (Unit 3)
Topic 3.12's payoff (LO 3.12.C) is using equivalent forms to solve equations. Choosing the right face of cos(2θ) can turn a mixed sine-and-cosine equation into a single-function quadratic you can factor and solve on the unit circle.
Sinusoidal functions and the unit circle (Unit 3)
Double-angle identities explain graph behavior too. sin(2θ) = 2 sin θ cos θ shows why doubling the input doesn't double the output, which connects to period changes when you saw f(θ) = sin(bθ) earlier in Unit 3.
This shows up almost entirely in the non-calculator multiple-choice style of question. Typical stems ask which expression is NOT equivalent to cos(2θ), or which identity you should apply to rewrite cos(2α) using only sine and cosine of α. The trap answers are always the fake-equivalent forms, like 2cos²θ + 1 instead of 2cos²θ − 1, or sin²θ − cos²θ with the subtraction flipped. You also need double-angle substitutions when solving trig equations, where rewriting cos(2θ) in terms of just sin θ or just cos θ turns the problem into a factorable quadratic. No released FRQ has demanded the term by name, but the rewrite-then-solve skill it builds is exactly what Topic 3.12 equation-solving questions reward.
Sum identities handle two different angles, like sin(α + β), while double-angle identities handle one angle doubled, like sin(2θ). They're not separate formula families. The double-angle versions fall straight out of the sum identities when you set α = β. So sin(60° − 45°) calls for a difference identity, but sin(2 · 30°) calls for the double-angle form, and both trace back to the same source. The other classic mix-up is thinking cos(2θ) = 2cos θ. It doesn't. Cosine isn't linear, and that wrong answer is planted on nearly every quiz.
The double-angle identities are sin(2θ) = 2 sin θ cos θ and cos(2θ) = cos²θ − sin²θ, and both come from setting α = β in the sum identities.
cos(2θ) has three equivalent forms (cos²θ − sin²θ, 2cos²θ − 1, and 1 − 2sin²θ) because the Pythagorean identity lets you swap sin²θ and cos²θ for each other.
Choose the form of cos(2θ) that matches the equation you're solving. If the rest of the equation has sin θ, use 1 − 2sin²θ so everything is in one function.
cos(2θ) is NOT equal to 2cos θ, and sin(2θ) is NOT equal to 2sin θ. You cannot pull a constant out of a trig function.
On multiple choice, expect to identify which expression is or isn't equivalent to sin(2θ) or cos(2θ), so verify candidates by deriving from the sum identities instead of guessing.
They are sin(2θ) = 2 sin θ cos θ and cos(2θ) = cos²θ − sin²θ, where the cosine version can also be written as 2cos²θ − 1 or 1 − 2sin²θ. They're part of Topic 3.12 in Unit 3 and support LO 3.12.B.
No. Cosine isn't a linear function, so you can't factor the 2 out. cos(2θ) equals cos²θ − sin²θ, and you can confirm with numbers, since cos(120°) = −0.5 but 2cos(60°) = 1.
A sum identity expands a function of two different angles, like cos(α + β) = cos α cos β − sin α sin β. A double-angle identity is just that same formula with both angles equal, so cos(2θ) = cos(θ + θ) = cos²θ − sin²θ. Knowing the sum identities means you get the double-angle ones for free.
You should know all three exist, but you only need to memorize cos²θ − sin²θ. The other two forms (2cos²θ − 1 and 1 − 2sin²θ) come from substituting the Pythagorean identity, which is a quick one-step rewrite under LO 3.12.A.
Because it converts a mixed equation into a single-function one. For example, in cos(2θ) = sin θ, replacing cos(2θ) with 1 − 2sin²θ gives 2sin²θ + sin θ − 1 = 0, a quadratic in sin θ that factors. That's exactly the skill LO 3.12.C describes.
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