Sin A Sin B Formula: Product-to-Sum Proof & Examples

The Trick That Multiplied Numbers Before Logarithms Existed

Before logarithms existed, astronomers multiplied enormous numbers by turning them into sines and adding them. The method was called prosthaphaeresis, and for about fifty years it was the fastest way on Earth to multiply - Tycho Brahe's observatory ran on it. The whole trick rests on one idea: a product of two trig values can be rewritten as a sum, and adding is far easier than multiplying by hand.

That idea is exactly the sin A sin B formula. It takes the product $\sin A \sin B$ and converts it into a difference of cosines you can read straight off a table. The same conversion is why the identity still earns its place today in calculus, wave physics, and signal processing - anywhere a product of oscillations needs to become a sum you can integrate or measure.

What Is The Sin A Sin B Formula?

The sin A sin B formula is a product-to-sum identity. It states that the product of the sines of two angles equals half the difference of two cosines:

$$\sin A \sin B = \frac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]$$

Here $A$ and $B$ are any two angles, in degrees or radians. The right side uses the difference angle $A-B$ and the sum angle $A+B$. Because both cosines can be looked up or computed independently, the formula turns a single multiplication into a subtraction of two known values.

A quick word on the name. A product-to-sum identity starts with a product (two functions multiplied) and ends with a sum or difference (functions added or subtracted). It is the mirror image of a sum-to-product formula, which runs the other way. Do not confuse this with sin A + sin B: that one adds two sines, this one multiplies them, and the two identities are not interchangeable. The full family lives among the standard trigonometric identities.

Reading The Formula Both Ways

The identity is genuinely two-directional, and knowing that is what makes it useful:

• Left to right (product to sum): you have $\sin A \sin B$ and want a sum - the form calculus and integration prefer.

• Right to left (sum to product): you have a difference of cosines and want a product - the form that helps you factor and simplify.

What Are The Properties Of The Sin A Sin B Formula?

A few properties fall straight out of the formula and are worth holding onto.

• Symmetry in $A$ and $B$. Swapping $A$ and $B$ leaves the product $\sin A \sin B$ unchanged, and the formula respects that: $\cos(A-B) = \cos(B-A)$ because cosine is even, and $\cos(A+B)$ is already symmetric. So $\sin A \sin B = \sin B \sin A$, as it must.

• The output is a difference, not a sum. For the sine product you subtract: $\cos(A-B) - \cos(A+B)$. The cos A cos B product adds the same two cosines instead. Getting the sign wrong here is the single most common error, and the next section shows why the minus belongs.

• Equal angles give a power-reduction identity. Set $B = A$. Then $\sin A \sin A = \sin^2 A = \frac{1}{2}\left[\cos 0 - \cos 2A\right] = \frac{1 - \cos 2A}{2}$. That is the standard half-angle / power-reduction result, and it is just the sin A sin B formula with both angles the same.

• Range check. Since each cosine sits in $[-1, 1]$, the bracket $\cos(A-B) - \cos(A+B)$ sits in $[-2, 2]$, so the product $\sin A \sin B$ stays in $[-1, 1]$ - exactly the range a product of two sines should have.

How Do You Derive The Sin A Sin B Formula?

The proof needs only the two cosine angle formulas, which come from the sum and difference identities. Write them out:

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

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

Look at the two right-hand sides. They are identical except for the sign in front of $\sin A \sin B$. Subtract the second equation from the first, and the $\cos A \cos B$ terms cancel:

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

Divide both sides by $2$:

$$\sin A \sin B = \frac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]$$

That is the whole derivation. The reason the answer is a difference of cosines is baked into the algebra: subtracting the two formulas is what isolates $\sin A \sin B$, and subtraction is what leaves the minus sign between the cosines. If you had instead wanted $\cos A \cos B$, you would have added the two formulas - which is why that identity comes out as a sum.

Where Is The Sin A Sin B Formula Used? - "Turning products into sums you can add"

The formula exists to solve one problem: a product of oscillations is hard to work with, and a sum is easy. That single reframing is why it keeps showing up.

• Integration in calculus. You cannot integrate $\sin 2x \sin 5x$ by inspection, but you can integrate $\frac{1}{2}[\cos 3x - \cos 7x]$ term by term in seconds. Every calculus course reaches for the product-to-sum identities the moment integrals of products appear.

• Wave and sound physics. When two sound waves of nearby frequency overlap, their product produces the pulsing "beat" you hear. The sin A sin B formula rewrites that product as two separate cosine waves - one slow, one fast - which is precisely the slow beat and the fast tone your ear picks apart.

• The historical origin. The prosthaphaeresis method used this identity to replace multiplication with addition in 16th-century astronomy and navigation. Getting a position wrong at sea was a survival matter, and this identity was the shortcut that made the arithmetic fast enough to be practical.

What competitor explainers usually skip is why the product becomes a difference rather than a sum. It is not arbitrary - it is a direct fingerprint of the subtraction step in the proof, and once you have seen that step, you never again guess the sign.

Examples Of The Sin A Sin B Formula

Example 1

Rewrite $\sin 5x \sin 3x$ as a sum or difference.

Apply the formula with $A = 5x$ and $B = 3x$:

$$\sin 5x \sin 3x = \frac{1}{2}\left[\cos(5x - 3x) - \cos(5x + 3x)\right] = \frac{1}{2}\left[\cos 2x - \cos 8x\right]$$

Final answer: $\sin 5x \sin 3x = \frac{1}{2}\left[\cos 2x - \cos 8x\right]$.

Example 2

Evaluate $\sin 75^\circ \sin 15^\circ$. First instinct, then the correct route.

The tempting move is to reach for a product rule that adds the cosines: $\frac{1}{2}[\cos(A-B) + \cos(A+B)]$. Try it and you would get $\frac{1}{2}[\cos 60^\circ + \cos 90^\circ] = \frac{1}{2}\left[\frac{1}{2} + 0\right] = \frac{1}{4}$.

Check that against a rough estimate. Both $\sin 75^\circ$ and $\sin 15^\circ$ are positive and their product should be modest and positive, so the size feels plausible — but the identity you used is the one for $\cos A \cos B$, not $\sin A \sin B$. The sine product subtracts.

Use the correct form:

$$\sin 75^\circ \sin 15^\circ = \frac{1}{2}\left[\cos(75^\circ - 15^\circ) - \cos(75^\circ + 15^\circ)\right] = \frac{1}{2}\left[\cos 60^\circ - \cos 90^\circ\right]$$

$$= \frac{1}{2}\left[\frac{1}{2} - 0\right] = \frac{1}{4}$$

Here the two answers happen to coincide because $\cos 90^\circ = 0$, but that is luck, not licence — pick a case where the extra cosine is nonzero and the wrong sign gives a wrong number.

Final answer: $\sin 75^\circ \sin 15^\circ = \dfrac{1}{4}$.

Example 3

Find the exact value of $\sin 45^\circ \sin 15^\circ$.

With $A = 45^\circ$ and $B = 15^\circ$:

$$\sin 45^\circ \sin 15^\circ = \frac{1}{2}\left[\cos 30^\circ - \cos 60^\circ\right] = \frac{1}{2}\left[\frac{\sqrt{3}}{2} - \frac{1}{2}\right]$$

$$= \frac{\sqrt{3} - 1}{4}$$

Final answer: $\sin 45^\circ \sin 15^\circ = \dfrac{\sqrt{3} - 1}{4}$.

Example 4

Use the formula to evaluate the integral $\displaystyle\int \sin 6x \sin 2x , dx$.

Convert the product first:

$$\sin 6x \sin 2x = \frac{1}{2}\left[\cos 4x - \cos 8x\right]$$

Now integrate term by term:

$$\int \frac{1}{2}\left[\cos 4x - \cos 8x\right] dx = \frac{1}{2}\left[\frac{\sin 4x}{4} - \frac{\sin 8x}{8}\right] + C$$

Final answer: $\dfrac{\sin 4x}{8} - \dfrac{\sin 8x}{16} + C$.

Example 5

Show that $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$ using the sin A sin B formula.

Set $A = B = \theta$, so the product is $\sin \theta \sin \theta = \sin^2 \theta$:

$$\sin^2 \theta = \frac{1}{2}\left[\cos(\theta - \theta) - \cos(\theta + \theta)\right] = \frac{1}{2}\left[\cos 0 - \cos 2\theta\right]$$

Since $\cos 0 = 1$:

$$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$$

Final answer: the power-reduction identity is just the sin A sin B formula with equal angles.

Example 6

Simplify $2\sin 3x \sin x$ and confirm the leading coefficient disappears.

The stray $2$ out front cancels the $\frac{1}{2}$ inside the formula:

$$2\sin 3x \sin x = 2 \times \frac{1}{2}\left[\cos(3x - x) - \cos(3x + x)\right] = \cos 2x - \cos 4x$$

Final answer: $2\sin 3x \sin x = \cos 2x - \cos 4x$.

The first-instinct error students reach for across these is grabbing the addition version of the product rule out of habit - the sine product is the one that subtracts, and mixing it up with the cosine product is exactly where the sign flips.

Common Mistakes With The Sin A Sin B Formula

Mistake 1: Using A Plus Sign Instead Of A Minus

Where it slips in: Any time you write the product-to-sum form from memory and blur it with the $\cos A \cos B$ identity.

Don't do this: Writing $\sin A \sin B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$.

The correct way: The sine product subtracts: $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$. The memorizer who learned "product equals half the sum of cosines" gets this right for $\cos A \cos B$ and wrong for $\sin A \sin B$, because they never traced which of the two proofs used subtraction.

Mistake 2: Swapping The Sum And Difference Angles

Where it slips in: Deciding which angle goes with which cosine.

Don't do this: Writing $\frac{1}{2}[\cos(A+B) - \cos(A-B)]$, which flips the whole sign.

The correct way: The difference angle $A-B$ pairs with the positive cosine, the sum angle $A+B$ with the negative one. The rusher who writes the two cosines in whatever order they come to mind ends up with the negative of the right answer. Anchor it to the proof: you subtracted $\cos(A+B)$ from $\cos(A-B)$, so $\cos(A-B)$ leads.

Mistake 3: Confusing The Product With Sin A Plus Sin B

Where it slips in: Reading a problem quickly and treating $\sin A \sin B$ as $\sin A + \sin B$.

Don't do this: Applying the sum-to-product formula $2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$ to a product.

The correct way: A product and a sum are different starting points with different identities. Read the operation between the two sines before you pick a formula.

Key Takeaways

• The sin A sin B formula is $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$, a product-to-sum identity.

• It is proved by subtracting the $\cos(A+B)$ formula from the $\cos(A-B)$ formula, which cancels $\cos A \cos B$.

• The output is a difference of cosines; the cosine product $\cos A \cos B$ is the one that adds.

• Setting $B = A$ turns it into the power-reduction identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$.

• Its main jobs are integrating products of sines and separating overlapping waves.

To take these identities further with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for identity practice, or live math classes online with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: rewrite $\sin 4x \sin 6x$ as a difference of cosines, then evaluate $\sin 105^\circ \sin 15^\circ$ using the formula. If you get stuck on the sign, come back to the derivation above and follow the subtraction step. Want a live Bhanzu trainer to work these through with you? Book a free demo class.

Read More

• Cos2x — the double-angle identity that power-reduction connects to.

• Sin Cos Tan — the three primary ratios these identities are built from.

• Trigonometric Table — the special-angle values used in every worked example.

• Trigonometric Functions — how sine, cosine, and tangent behave as functions.

• Pythagorean Identities — the other backbone identities of trigonometry.