Tutorials › Calculus II › Deriving Product-to-Sum Identities

Sine and Cosine Powers · Tutorial 112 of 998

Deriving Product-to-Sum Identities

Use angle-addition identities to derive three product-to-sum formulas and recognize when they make a trigonometric integral easier.

Beginner 10 min read

What You'll Learn

  • Derive the sine-sine product-to-sum identity by subtracting cosine angle-addition formulas.
  • Derive the cosine-cosine identity by adding those formulas.
  • Derive the sine-cosine identity from the sine addition and subtraction formulas.
  • Keep track of the signs and the order of the angles in each identity.
  • Recognize how product-to-sum formulas simplify products in integrals.

Why Convert a Product into a Sum?

A product such as \(\sin A\cos B\) can be awkward to integrate directly, especially when the angles \(A\) and \(B\) differ. A product-to-sum identity rewrites it as a sum or difference of individual sine or cosine terms. Those terms are often easier to handle because their antiderivatives are standard.

The identities are also connected to power-reduction formulas. Setting the two angles equal turns products such as \(\sin^2 x\) and \(\cos^2 x\) into expressions involving a constant and a double angle. The main goal here is to derive these identities carefully and understand what each one does.

Definition: A product-to-sum identity is a trigonometric identity that expresses a product of sine and/or cosine functions as a sum or difference of individual trigonometric functions.

The Three Identities

For any real angles \(A\) and \(B\), the product-to-sum formulas are:

$$ \sin A\sin B = \frac{1}{2}\bigl(\cos(A-B)-\cos(A+B)\bigr) $$
$$ \cos A\cos B = \frac{1}{2}\bigl(\cos(A-B)+\cos(A+B)\bigr) $$
$$ \sin A\cos B = \frac{1}{2}\bigl(\sin(A+B)+\sin(A-B)\bigr) $$

The order of the angles matters in the last formula: it is written for \(\sin A\cos B\). For the reversed product, the corresponding identity is

$$ \cos A\sin B = \frac{1}{2}\bigl(\sin(A+B)-\sin(A-B)\bigr). $$

These are identities, not integration rules that apply only in a particular setting. They hold for every pair of real angles. In an integral, however, their usefulness depends on whether the resulting terms are simpler to integrate than the original product.

Derivation from Angle-Addition Formulas

The cosine addition and subtraction formulas are

$$ \cos(A+B)=\cos A\cos B-\sin A\sin B, \qquad \cos(A-B)=\cos A\cos B+\sin A\sin B. $$

Subtracting the first equation from the second cancels the \(\cos A\cos B\) terms and leaves \(2\sin A\sin B\). Adding them cancels the sine products and leaves \(2\cos A\cos B\). The sine-product and cosine-product identities therefore follow directly.

1
Subtract the cosine formulas.
\(\cos(A-B)-\cos(A+B)=2\sin A\sin B\). Dividing by \(2\) gives the sine-sine identity.
2
Add the cosine formulas.
\(\cos(A-B)+\cos(A+B)=2\cos A\cos B\). Dividing by \(2\) gives the cosine-cosine identity.
3
Add the sine formulas.
The sine addition and subtraction formulas isolate the sine-cosine product, as shown below.

For the remaining identity, use the sine addition and subtraction formulas:

$$ \sin(A+B)=\sin A\cos B+\cos A\sin B, $$
$$ \sin(A-B)=\sin A\cos B-\cos A\sin B. $$

Adding these equations cancels the \(\cos A\sin B\) terms. The left side is the sum of the two sine terms, and the right side is \(2\sin A\cos B\). Dividing by \(2\) proves the identity.

Why the signs differ: For \(\sin A\sin B\), subtracting the cosine formulas isolates the product; for \(\cos A\cos B\), adding them does. For \(\sin A\cos B\), adding the sine formulas isolates the product.

Using the Identities When the Angles Match

When \(A=B\), the sine-sine identity gives a constant minus a cosine term, and the cosine-cosine identity gives a constant plus a cosine term. These are the power-reduction identities for the squares. When both factors have the same angle, the sine-cosine identity gives a single sine term with a doubled angle. This makes each product straightforward to integrate using basic antiderivatives.

Worked Example: Integrating a Sine-Squared Product

Rewrite and integrate \(\sin^2(3x)\). Set \(A=B=3x\) in the sine-sine identity:

$$ \sin^2(3x) = \frac{1}{2}\bigl(\cos(3x-3x)-\cos(3x+3x)\bigr) = \frac{1}{2}\bigl(\cos 0-\cos 6x\bigr). $$

Since \(\cos 0=1\), this becomes

$$ \sin^2(3x)=\frac{1}{2}-\frac{1}{2}\cos 6x. $$

Integrate each term:

$$ \int \sin^2(3x)\,dx = \frac{x}{2}-\frac{1}{2}\cdot\frac{\sin 6x}{6}+C = \frac{x}{2}-\frac{\sin 6x}{12}+C. $$

To check, differentiate the result:

$$ \frac{d}{dx}\left(\frac{x}{2}-\frac{\sin 6x}{12}\right) = \frac{1}{2}-\frac{6\cos 6x}{12} = \frac{1}{2}-\frac{1}{2}\cos 6x = \sin^2(3x). $$

The derivative matches the integrand.

Worked Example: Integrating a Cosine-Squared Product

Rewrite and integrate \(\cos^2(2x)\). Set \(A=B=2x\) in the cosine-cosine identity:

$$ \cos^2(2x) = \frac{1}{2}\bigl(\cos(2x-2x)+\cos(2x+2x)\bigr) = \frac{1}{2}\bigl(\cos 0+\cos 4x\bigr). $$

Using \(\cos 0=1\), obtain

$$ \cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos 4x. $$

Now integrate:

$$ \int \cos^2(2x)\,dx = \frac{x}{2}+\frac{1}{2}\cdot\frac{\sin 4x}{4}+C = \frac{x}{2}+\frac{\sin 4x}{8}+C. $$

Differentiate to verify:

$$ \frac{d}{dx}\left(\frac{x}{2}+\frac{\sin 4x}{8}\right) = \frac{1}{2}+\frac{4\cos 4x}{8} = \frac{1}{2}+\frac{1}{2}\cos 4x = \cos^2(2x). $$

The derivative recovers the original integrand.

Worked Example: Integrating a Sine-Cosine Product

Use the sine-cosine identity to integrate \(\sin(4x)\cos(4x)\). With \(A=B=4x\),

$$ \sin(4x)\cos(4x) = \frac{1}{2}\bigl(\sin(4x+4x)+\sin(4x-4x)\bigr) = \frac{1}{2}\bigl(\sin 8x+\sin 0\bigr) = \frac{1}{2}\sin 8x. $$

Since \(\sin 0=0\), integrate the resulting single sine term:

$$ \int \sin(4x)\cos(4x)\,dx = \frac{1}{2}\int \sin 8x\,dx = -\frac{\cos 8x}{16}+C. $$

Check by differentiation:

$$ \frac{d}{dx}\left(-\frac{\cos 8x}{16}\right) = \frac{8\sin 8x}{16} = \frac{1}{2}\sin 8x = \sin(4x)\cos(4x). $$

Thus the antiderivative differentiates to the stated product.

Recognizing When a Product-to-Sum Formula Helps

Each formula changes a particular kind of product into a sum of one-family terms. The choice is determined by the two factors:

ProductResult of the identityUseful when
\(\sin A\sin B\)A difference of cosinesIntegrating the resulting cosine terms is more direct than handling the product.
\(\cos A\cos B\)A sum of cosinesThe individual cosine terms have simpler antiderivatives than the original product.
\(\sin A\cos B\)A sum of sinesThe sine terms are easier to integrate separately than the product.

If the angles match, one of the resulting angles is zero. The corresponding cosine term is the constant \(1\), or the corresponding sine term is \(0\). That is why the formulas simplify squares and same-angle sine-cosine products so effectively.

If the angles do not match, the formulas instead produce terms involving the sum and difference of the angles. This can still be helpful: a product becomes a pair of individual trigonometric terms, each of which may be integrated separately. Before applying the identity, check that the resulting angle expressions are simpler for the integral at hand. A product-to-sum rewrite is not automatically an improvement if it creates terms that are no easier to work with.

A Common Sign Error

The sine-sine formula requires a difference of cosines in the order \(\cos(A-B)-\cos(A+B)\). Reversing the sign and adding the two cosine expressions gives the cosine-cosine product instead. For example, the incorrect claim

$$ \cos(A-B)+\cos(A+B)=2\sin A\sin B $$

fails when \(A=B=\frac{\pi}{3}\). Its left side is

$$ \cos 0+\cos\left(\frac{2\pi}{3}\right) = 1-\frac{1}{2} = \frac{1}{2}, $$

whereas its right side is

$$ 2\sin^2\left(\frac{\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{2}. $$

The values are different, so the claimed identity is false. The correction is to subtract the cosine formulas to isolate the sine product:

$$ \cos(A-B)-\cos(A+B)=2\sin A\sin B. $$

A reliable way to avoid this mistake is to write both angle-addition equations before adding or subtracting them. The terms that should cancel then determine the correct operation.

Check Your Understanding

Use the angle-addition formulas and the product-to-sum identities to answer these questions.

  1. Which operation on the two cosine angle-addition formulas gives the sine-sine product identity?
  2. Write \(\cos A\cos B\) as a sum of cosines.
  3. Use the sine-cosine identity to rewrite \(\sin(2x)\cos(2x)\).
  4. Why does setting \(A=B\) in the sine-sine identity introduce a constant term?
  5. What should you check before deciding that a product-to-sum rewrite will simplify an integral?

Test yourself: Calculus II final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon