2sinAsinB Formula: Proof, Examples & Uses

#Trigonometry
TL;DR
The 2sinAsinB formula is a product-to-sum identity: 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B). It turns a product of two sines into a difference of two cosines, and it is the one product-to-sum rule that carries a minus sign. For a quick check, with A=60∘A = 60^\circ and B=30∘B = 30^\circ both sides equal 32≈0.8660\frac{\sqrt{3}}{2} \approx 0.8660.
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Bhanzu TeamLast updated on September 10, 202610 min read

What Is The 2sinAsinB Formula?

The 2sinAsinB formula states that 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B). It is a product-to-sum identity, meaning it rewrites a product of two trigonometric ratios as a sum or difference of others. Here the product of two sines becomes the difference of two cosines.

Two features make this identity worth memorising on its own:

  • It produces cosines, not sines. The inputs are two sines, but the output is built entirely from cosine terms.

  • It is the only product-to-sum rule with a minus sign. Its close cousin 2cos⁡Acos⁡B2\cos A\cos B adds the two cosines; the sine-sine version subtracts them.

Dividing both sides by two gives the half-angle-free companion form, sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\sin B = \tfrac{1}{2}\left[\cos(A-B) - \cos(A+B)\right]. The two forms are the same identity; use whichever leaves the cleaner arithmetic. The whole family lives inside the trigonometric identities toolkit, and it is built directly from the sum and difference identities.

How Do You Derive The 2sinAsinB Formula?

The derivation needs only the two cosine angle formulas, one for a difference of angles and one for a sum. Write them side by side:

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A\cos B + \sin A\sin B

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B

The two right-hand sides share the term cos⁡Acos⁡B\cos A\cos B. That shared term is what cancels. Subtract the second equation from the first:

cos⁡(A−B)−cos⁡(A+B)=(cos⁡Acos⁡B+sin⁡Asin⁡B)−(cos⁡Acos⁡B−sin⁡Asin⁡B)\cos(A - B) - \cos(A + B) = \left(\cos A\cos B + \sin A\sin B\right) - \left(\cos A\cos B - \sin A\sin B\right)

Remove the brackets, watching the signs on the second line:

cos⁡(A−B)−cos⁡(A+B)=cos⁡Acos⁡B+sin⁡Asin⁡B−cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) - \cos(A + B) = \cos A\cos B + \sin A\sin B - \cos A\cos B + \sin A\sin B

The cos⁡Acos⁡B\cos A\cos B terms cancel, and the two sin⁡Asin⁡B\sin A\sin B terms add:

cos⁡(A−B)−cos⁡(A+B)=2sin⁡Asin⁡B\cos(A - B) - \cos(A + B) = 2\sin A\sin B

Flip the equation to read it the usual way:

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B)

That is the whole proof. The minus sign in the identity is a direct inheritance of the minus sign in the cos⁡(A+B)\cos(A+B) expansion, which is the deep reason the sine-sine rule subtracts while the cosine-cosine rule adds.

What Is A Worked Example Of 2sinAsinB?

The fastest way to trust the identity is to compute both sides and watch them match. All three examples below show the angle in degrees, with the radian form given where a special angle appears.

Example 1: A clean numeric check with A=60∘A = 60^\circ and B=30∘B = 30^\circ.

Left-hand side, using sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2} and sin⁡30∘=12\sin 30^\circ = \frac{1}{2}:

2sin⁡60∘sin⁡30∘=2⋅32⋅12=32≈0.86602\sin 60^\circ\sin 30^\circ = 2\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2} = \frac{\sqrt{3}}{2} \approx 0.8660

Right-hand side, with A−B=30∘=π6A - B = 30^\circ = \frac{\pi}{6} and A+B=90∘=π2A + B = 90^\circ = \frac{\pi}{2}:

cos⁡30∘−cos⁡90∘=32−0=32≈0.8660\cos 30^\circ - \cos 90^\circ = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \approx 0.8660

Both sides equal 32\frac{\sqrt{3}}{2}, so the identity holds. You can cross-check the individual values against the trigonometric table or the sin 60 degrees page.

Example 2: A product that looks hard but collapses, 2sin⁡75∘sin⁡15∘2\sin 75^\circ\sin 15^\circ.

Multiplying sin⁡75∘\sin 75^\circ by sin⁡15∘\sin 15^\circ directly is awkward. The identity turns it into a difference of familiar cosines, with A−B=60∘A - B = 60^\circ and A+B=90∘A + B = 90^\circ:

2sin⁡75∘sin⁡15∘=cos⁡60∘−cos⁡90∘=12−0=122\sin 75^\circ\sin 15^\circ = \cos 60^\circ - \cos 90^\circ = \frac{1}{2} - 0 = \frac{1}{2}

A decimal check confirms it: 2×0.96593×0.25882≈0.50002 \times 0.96593 \times 0.25882 \approx 0.5000.

Example 3: Rewriting a product of sines of multiples, 2sin⁡5xsin⁡3x2\sin 5x\sin 3x.

The identity works with variable angles too. Set A=5xA = 5x and B=3xB = 3x, so A−B=2xA - B = 2x and A+B=8xA + B = 8x:

2sin⁡5xsin⁡3x=cos⁡2x−cos⁡8x2\sin 5x\sin 3x = \cos 2x - \cos 8x

This is exactly the step that makes a product of sines integrable, since a difference of cosines integrates term by term while a raw product does not.

What Are The Sibling Product-To-Sum Formulas?

The 2sinAsinB formula is one of four product-to-sum identities. They all come from adding or subtracting the sum and difference expansions of sine and cosine, and they are easiest to remember as a set.

Table: The four product-to-sum identities and what each one produces.

Product

Equals

Output

2sin⁡Asin⁡B2\sin A\sin B

cos⁡(A−B)−cos⁡(A+B)\cos(A-B) - \cos(A+B)

cosines, subtracted

2cos⁡Acos⁡B2\cos A\cos B

cos⁡(A−B)+cos⁡(A+B)\cos(A-B) + \cos(A+B)

cosines, added

2sin⁡Acos⁡B2\sin A\cos B

sin⁡(A+B)+sin⁡(A−B)\sin(A+B) + \sin(A-B)

sines, added

2cos⁡Asin⁡B2\cos A\sin B

sin⁡(A+B)−sin⁡(A−B)\sin(A+B) - \sin(A-B)

sines, added or subtracted

Two patterns keep them straight. When both factors are cosines or both are sines, the answer is made of cosines; when one factor is a sine and one is a cosine, the answer is made of sines. And only the sine-sine row subtracts, which is the single fact most worth locking in. Running these in reverse gives the sum to product formulas, the mirror-image toolkit for turning sums back into products.

Why Does 2sinAsinB Equal cos(A−B) − cos(A+B)?

The identity is not an arbitrary rule to memorise. It falls out of one honest fact about the cosine of a sum, and it has a clean picture on the unit circle.

  • The minus sign is inherited. The expansion cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin B already carries a minus in front of sin⁡Asin⁡B\sin A\sin B. Subtracting that expansion flips the minus to a plus and doubles the sine product, which is where the leading 2 comes from.

  • The cosines survive because their shared part cancels. Both cos⁡(A±B)\cos(A\pm B) expansions begin with cos⁡Acos⁡B\cos A\cos B. Subtracting removes that common piece and leaves only the sine-sine terms, so the product of sines is what remains.

  • It is a distance on the unit circle. cos⁡(A−B)\cos(A-B) and cos⁡(A+B)\cos(A+B) are the x-coordinates of two points on the unit circle. Their difference, cos⁡(A−B)−cos⁡(A+B)\cos(A-B) - \cos(A+B), is the horizontal gap between those points, and that gap is exactly 2sin⁡Asin⁡B2\sin A\sin B.

There is a bonus that appears when the two angles are equal. Setting B=AB = A gives 2sin⁡Asin⁡A=cos⁡0−cos⁡2A2\sin A\sin A = \cos 0 - \cos 2A, which simplifies to the power-reduction identity 2sin⁡2A=1−cos⁡2A2\sin^{2}A = 1 - \cos 2A. The same subtraction that handles two different angles also lowers the power of a squared sine, which is why this identity shows up constantly in calculus.

Who Discovered The Product-To-Sum Formulas?

Product-to-sum thinking is older than the algebra we use to write it. For centuries its main job was not simplification but survival, since it was how astronomers multiplied enormous numbers before anyone had a better tool.

Two other names sit on either side of that story:

  • Ibn Yunus (c. 950–1009, Egypt) was an astronomer in Cairo whose work on trigonometric relations is often cited as an early source for the product-to-sum idea, generations before European observatories put it to work.

  • John Napier (1550–1617, Scotland) published his logarithms in 1614, and they were so much faster than prosthaphaeresis that the older method quietly disappeared. The product-to-sum formulas stayed, but their job shifted from multiplying to simplifying, which is how students meet them today. Earlier still, Ptolemy's Almagest (c. 150 CE) had tabulated chords, the ancestor of the sine tables all of this depended on.

Where Is 2sinAsinB Used In The Real World?

The identity earns its keep wherever two oscillations multiply and the result needs to be read as separate frequencies.

  • Acoustics and musical beats: two tones of nearly equal pitch played together produce a slow throb, the "beat." Multiplying their two sine waves and applying the 2sinAsinB formula shows the pulse as a low difference frequency, which is exactly what a piano tuner listens for.

  • Fourier analysis: breaking a signal into pure sine and cosine components relies on integrating products of trigonometric functions, and product-to-sum identities are what make those integrals solvable.

  • Signal processing and radio: mixing two signals multiplies them, and the formula predicts the sum and difference frequencies that appear, the basis of how a radio shifts a station up or down the dial.

  • Calculus and physics: any integral of a product of sines, from wave interference to alternating current, is first rewritten as a difference of cosines using this exact identity.

One subtraction connects a piano tuner's ear, a radio dial, and a physics exam. That reach is the reason product-to-sum identities survived long after the calculation they were invented for.

What Are The Most Common Mistakes With 2sinAsinB?

Three errors account for most lost marks on this identity. Each one is a small slip with a fixed cure.

Reversing the order of subtraction.

Where it slips in:

A student writes 2sin⁡Asin⁡B=cos⁡(A+B)−cos⁡(A−B)2\sin A\sin B = \cos(A+B) - \cos(A-B), subtracting the smaller-angle cosine from the larger.

Don't do this:

Do not lead with cos⁡(A+B)\cos(A+B). Reversing the order negates the whole answer, turning a correct value into its opposite.

The correct way:

Always write cos⁡(A−B)\cos(A-B) first: 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B). The difference angle comes first, then the sum angle is subtracted.

Confusing it with the 2cosAcosB formula.

Where it slips in:

A student remembers that the cosine terms appear but adds them, writing cos⁡(A−B)+cos⁡(A+B)\cos(A-B) + \cos(A+B), which is actually the rule for 2cos⁡Acos⁡B2\cos A\cos B.

Don't do this:

Do not add the cosines for a product of sines. Adding gives the cosine-cosine identity, not the sine-sine one.

The correct way:

Tie the sign to the factors: sine times sine subtracts, cosine times cosine adds. For our target, 2sin⁡Asin⁡B2\sin A\sin B subtracts.

Dropping the factor of 2 or the compensating one-half.

Where it slips in:

A student uses the half-angle-free form sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\sin B = \tfrac{1}{2}\left[\cos(A-B) - \cos(A+B)\right] but forgets the 12\tfrac{1}{2}, or uses the doubled form and forgets the leading 2.

Don't do this:

Do not mix the two forms. Writing sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)\sin A\sin B = \cos(A-B) - \cos(A+B) without the half is off by a factor of two.

The correct way:

Keep the pair together. If the left side has 2sin⁡Asin⁡B2\sin A\sin B, the right side has no fraction; if the left side is sin⁡Asin⁡B\sin A\sin B, the right side carries the 12\tfrac{1}{2}.

Practice Problems On 2sinAsinB

Apply 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B) to each. Answers follow every line.

  1. Express 2sin⁡3xsin⁡x2\sin 3x\sin x as a difference of cosines.
    (Answer: cos⁡2x−cos⁡4x\cos 2x - \cos 4x.)

  2. Evaluate 2sin⁡75∘sin⁡15∘2\sin 75^\circ\sin 15^\circ.
    (Answer: cos⁡60∘−cos⁡90∘=12−0=12\cos 60^\circ - \cos 90^\circ = \frac{1}{2} - 0 = \frac{1}{2}.)

  3. Evaluate 2sin⁡60∘sin⁡30∘2\sin 60^\circ\sin 30^\circ.
    (Answer: cos⁡30∘−cos⁡90∘=32≈0.8660\cos 30^\circ - \cos 90^\circ = \frac{\sqrt{3}}{2} \approx 0.8660.)

  4. Write 2sin⁡7θsin⁡2θ2\sin 7\theta\sin 2\theta as a difference of cosines.
    (Answer: cos⁡5θ−cos⁡9θ\cos 5\theta - \cos 9\theta.)

  5. Use B=AB = A to simplify 2sin⁡2A2\sin^{2}A.
    (Answer: 2sin⁡Asin⁡A=cos⁡0−cos⁡2A=1−cos⁡2A2\sin A\sin A = \cos 0 - \cos 2A = 1 - \cos 2A.)

  6. Express sin⁡4xsin⁡6x\sin 4x\sin 6x as one-half of a difference.
    (Answer: 12(cos⁡2x−cos⁡10x)\tfrac{1}{2}\left(\cos 2x - \cos 10x\right).)

Where Should You Go Next After 2sinAsinB?

The 2sinAsinB formula is one door into the wider identity toolkit, and a few natural next steps open from here.

  1. Sum and difference identities. The two cosine expansions that this proof rests on, plus the sine versions, are the foundation for every product-to-sum rule.

  2. Sum to product formulas. The reverse direction, turning a sum or difference of sines and cosines back into a product.

  3. Cos A minus B formula. A close look at the single expansion that supplies the plus sign in this derivation.

If your child is building fluency with identities like this, a live Bhanzu trainer teaches product-to-sum reasoning from the "why" (the cancelling cosine terms, not just the memorised rule) in the Bhanzu trigonometry program.

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Frequently Asked Questions

What is the 2sinAsinB formula?
The 2sinAsinB formula is the product-to-sum identity 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B). It rewrites a product of two sines as the difference of two cosines, and its half form is sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A\sin B = \tfrac{1}{2}\left[\cos(A-B) - \cos(A+B)\right].
Why is there a minus sign in the 2sinAsinB formula?
The minus sign comes straight from the expansion cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin B. Subtracting that expansion from the one for cos⁡(A−B)\cos(A-B) cancels the shared cosine term and doubles the sine product, which is why the sine-sine rule subtracts while the cosine-cosine rule adds.
How is sin A sin B different from 2 sin A cos B?
A product of two sines converts to a difference of cosines, while a product of a sine and a cosine converts to a sum of sines: 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B). The type of output, cosines versus sines, is set by the mix of factors you start with.
Can the 2sinAsinB formula give a negative value?
Yes. Because cos⁡(A−B)−cos⁡(A+B)\cos(A-B) - \cos(A+B) is a difference, the result is negative whenever cos⁡(A+B)\cos(A+B) exceeds cos⁡(A−B)\cos(A-B), which happens when one of the sine factors is itself negative. The identity holds for every pair of angles, positive, negative, or measured in radians.
What is 2sinAsinB when A equals B?
Setting B=AB = A gives 2sin⁡2A=1−cos⁡2A2\sin^{2}A = 1 - \cos 2A, the power-reduction identity. This is the special case that lowers a squared sine to a single cosine of the double angle, and it is one of the most-used steps in integration.
How do you use the formula in integration?
A raw product such as sin⁡5xsin⁡3x\sin 5x\sin 3x has no simple antiderivative, but rewriting it as 12(cos⁡2x−cos⁡8x)\tfrac{1}{2}(\cos 2x - \cos 8x) turns it into a difference of cosines that integrates term by term. Converting products to sums first is a standard opening move for trigonometric integrals.
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