AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & IdentitiesSum, Difference & Double-Angle Identities

Sum, Difference & Double-Angle Identities (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Sum identities

What are the sum identities for sine and cosine?

The sum identity for sine expresses $\sin (α + β)$ in terms of the sine and cosine of $α$ and $β$ separately
- $\sin (α + β) = \sin α \cos β + \cos α \sin β$

The sum identity for cosine expresses $\cos (α + β)$ similarly
- $\cos (α + β) = \cos α \cos β - \sin α \sin β$

These identities hold for all values of $α$ and $β$

How are sum identities used?

One common use is to find the exact value of a trigonometric function at a non-standard angle by writing that angle as a sum of two standard angles
- E.g. $\frac{7 π}{12} = \frac{π}{3} + \frac{π}{4}$
  - so $\sin \frac{7 π}{12}$ can be found using the sum identity with $α = \frac{π}{3}$ and $β = \frac{π}{4}$
Another common use is to rewrite trigonometric expressions in equivalent forms
- This can be helpful when solving equations or simplifying

Examiner Tips and Tricks

Note that the sum identities cannot be simplified by "distributing" the trigonometric function.

I.e, $\sin (α + β) \neq \sin α + \sin β$ in general

Difference identities

What are the difference identities?

The difference identities are obtained from the sum identities
- by replacing $β$ with $- β$
  - and using the fact that sine is odd ( $\sin (- β) = - \sin β$ )
  - and cosine is even ( $\cos (- β) = \cos β$ ):
- Doing this gives you
  - $\sin (α - β) = \sin α \cos β - \cos α \sin β$
  - $\cos (α - β) = \cos α \cos β + \sin α \sin β$

Notice the sign patterns
- The sine difference identity has a minus sign between its two terms
  - whereas the sine sum identity has a plus
- The cosine difference identity has a plus sign between its two terms
  - whereas the cosine sum identity has a minus

How are difference identities used?

Like sum identities, difference identities can be used to find exact values at non-standard angles by writing that angle as a difference of two standard angles
- E.g. $\frac{π}{12} = \frac{π}{3} - \frac{π}{4}$ , so $\sin \frac{π}{12}$ can be found using the difference identity

Examiner Tips and Tricks

The sign difference between the sum and difference identities for cosine is a common source of errors. A helpful way to remember:

Cosine formulas have opposite signs between the trig-function notation and the identity formula
- $\cos (α + β) = \cos α \cos β - \sin α \sin β$ (plus on the left, minus on the right)
- $\cos (α - β) = \cos α \cos β + \sin α \sin β$ (minus on the left, plus on the right)
Sine formulas have matching signs
- $\sin (α + β) = \sin α \cos β + \cos α \sin β$ (plus on the left, plus on the right)
- $\sin (α - β) = \sin α \cos β - \cos α \sin β$ (minus on the left, minus on the right)

Some people remember this as "cosine is contrary, sine is the same".

Worked Example

Find the exact value of each of the following expressions, using the sum or difference identities.

(a) $\sin \frac{π}{12}$

Answer:

Write $\frac{π}{12}$ as a difference of two standard angles

$\frac{π}{12} = \frac{π}{3} - \frac{π}{4}$

Apply the difference identity for sine

$\sin (\frac{π}{3} - \frac{π}{4}) = \sin \frac{π}{3} \cos \frac{π}{4} - \cos \frac{π}{3} \sin \frac{π}{4}$

Substitute in the exact values

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Therefore

$\sin \frac{π}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$

(b) $\cos \frac{7 π}{12}$

Answer:

Write $\frac{7 π}{12}$ as a sum of two standard angles

$\frac{7 π}{12} = \frac{π}{3} + \frac{π}{4}$

Apply the sum identity for cosine

$\cos (\frac{π}{3} + \frac{π}{4}) = \cos \frac{π}{3} \cos \frac{π}{4} - \sin \frac{π}{3} \sin \frac{π}{4}$

Substitute in the exact values

$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$

Therefore

$\cos \frac{7 π}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$

Double-angle identities

What are the double-angle identities?

The double-angle identities are obtained from the sum identities by setting $α = β = θ$ :
- For sine
  - $\sin (2 θ) = \sin (θ + θ) = \sin θ \cos θ + \cos θ \sin θ = 2 \sin θ \cos θ$
- For cosine
  - $\cos (2 θ) = \cos (θ + θ) = \cos θ \cos θ - \sin θ \sin θ = \cos^{2} θ - \sin^{2} θ$

This gives the two double-angle identities

$\sin (2 θ) = 2 \sin θ \cos θ$

$\cos (2 θ) = \cos^{2} θ - \sin^{2} θ$

What are the alternative forms of the cosine double-angle identity?

The cosine double-angle identity can be rewritten in two alternative forms by using the Pythagorean identity $\sin^{2} θ + \cos^{2} θ = 1$ :
- Substituting $\sin^{2} θ = 1 - \cos^{2} θ$ gives
  - $\cos (2 θ) = \cos^{2} θ - (1 - \cos^{2} θ) = 2 \cos^{2} θ - 1$
- Substituting $\cos^{2} θ = 1 - \sin^{2} θ$ gives
  - $\cos (2 θ) = (1 - \sin^{2} θ) - \sin^{2} θ = 1 - 2 \sin^{2} θ$

All three forms are equivalent

$\cos (2 θ) = \cos^{2} θ - \sin^{2} θ = 2 \cos^{2} θ - 1 = 1 - 2 \sin^{2} θ$

The different forms are useful in different situations
- e.g. if an expression already contains $\sin^{2} θ$ , then the $1 - 2 \sin^{2} θ$ form may be most convenient

How are double-angle identities used?

They are commonly used to rewrite expressions involving $\sin (2 x)$ or $\cos (2 x)$ in terms of $\sin x$ and $\cos x$
- or vice versa
They also arise when solving equations that mix single-angle and double-angle terms
- Applying the double-angle identity can convert the equation to one involving only single-angle terms

Worked Example

The function $g$ is given by $g (x) = 5 \cos (2 x)$ . Which of the following is an equivalent form for $g (x)$ ?

(A) $g (x) = 5 \cos^{2} x - 5 \sin^{2} x$

(B) $g (x) = 10 \cos x \sin x$

(D) $g (x) = 25 \cos x \sin x$

Answer

Apply the double-angle identity $\cos (2 x) = \cos^{2} x - \sin^{2} x$

$g (x) = 5 \cos (2 x) = 5 (\cos^{2} x - \sin^{2} x) = 5 \cos^{2} x - 5 \sin^{2} x$

That is option (A), but it's worth considering why the other options are incorrect

In option (B)
- $10 \cos x \sin x = 5 (2 \sin x \cos x) = 5 \sin 2 x$
In option (C)
- $5 - 10 \cos^{2} x = 5 (1 - 2 \cos^{2} x) = - 5 (2 \cos^{2} x - 1) = - 5 \cos 2 x$
In option (D)
- $25 \cos x \sin x = \frac{25}{2} (2 \sin x \cos x) = \frac{25}{2} \sin 2 x$

(A) $g (x) = 5 \cos^{2} x - 5 \sin^{2} x$

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Sum, Difference & Double-Angle Identities (College Board AP® Precalculus): Revision Note

Sum identities

What are the sum identities for sine and cosine?

How are sum identities used?

Examiner Tips and Tricks

Difference identities

What are the difference identities?

How are difference identities used?

Examiner Tips and Tricks

Worked Example

Double-angle identities

What are the double-angle identities?

What are the alternative forms of the cosine double-angle identity?

How are double-angle identities used?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis