Sums into Products

The compound angle page An angle written as the sum or difference of two others, such as A + B or A − B. The compound angle formulas express its sine, cosine and tangent in terms of the sines and cosines of A and B separately, which is necessary because none of the six functions is linear and none of them can be handed out over a sum. cos(A − B) = cos A cos B + sin A sin B is the one usually proved first; the other three follow from it by substituting −B for B and by the cofunction relationships. produced four expansions. Each one is two terms:

sin(A+B) = sin A cos B + cos A sin B

sin(A−B) = sin A cos B − cos A sin B

Write them one above the other and add. The two cos A sin B terms have opposite signs, so they cancel. The two sin A cos B terms are identical, so they double:

2 sin A cos B = sin(A+B) + sin(A−B)

Subtract the second line from the first instead. Now the sin A cos B terms cancel and the other pair doubles:

2 cos A sin B = sin(A+B) − sin(A−B)

Do the same with the two cosine expansions:

cos(A+B) = cos A cos B − sin A sin B

cos(A−B) = cos A cos B + sin A sin B

Adding cancels the sin A sin B pair:

2 cos A cos B = cos(A+B) + cos(A−B)

Subtracting cancels the cos A cos B pair and leaves cos(A+B) − cos(A−B) = −2 sin A sin B. Multiply both sides by minus 1 to get rid of the minus sign in front of the product:

2 sin A sin B = cos(A−B) − cos(A+B)

Those four are the product to sum The four identities that rewrite a product of two sines or cosines as a sum or difference of two: 2 sin A cos B = sin(A+B) + sin(A−B), 2 cos A sin B = sin(A+B) − sin(A−B), 2 cos A cos B = cos(A+B) + cos(A−B), and 2 sin A sin B = cos(A−B) − cos(A+B). Each is obtained by adding or subtracting two compound angle formulas, which cancels one of the two terms. Note the reversed order on the last one, which follows from cos(A+B) being the expansion that carries minus sin A sin B. formulas. Every one of them came from adding or subtracting two lines you already had.

The last one carries the trap. In the first three, the (A+B) term is written first. In the fourth, the difference runs the other way: it is cos(A−B) minus cos(A+B), not cos(A+B) minus cos(A−B).

The reason is visible in the expansions. cos(A+B) is the one with minus sin A sin B in it, so subtracting it is what makes the product come out positive. Writing it the other way round gives an answer with the wrong sign at every angle where the product is not zero.

Check the last one with numbers. Take A = 60 degrees and B = 30 degrees. Then 2 sin A sin B = 2 × (√3/2) × (1/2), which is √3/2. And cos(A−B) − cos(A+B) = cos(30 degrees) − cos(90 degrees) = √3/2 minus 0, which is √3/2. The two agree.

Written the other way round the right hand side would be minus √3/2, and the product of two positive numbers is not negative.

These formulas had a job before they had a name in a textbook. Multiplying two six figure numbers by hand is slow; adding them is fast. Astronomers used the product to sum formulas to convert a multiplication into a table lookup and an addition, a technique called prosthaphaeresis A method for multiplying large numbers using the product to sum identities, from the Greek for addition and subtraction. Looking up two cosines, adding them and halving is faster by hand than multiplying two six figure numbers, so astronomers used it to convert products into sums. Tycho Brahe's assistants were using it by about 1580, and it was the standard technique until Napier published logarithms in 1614.. Tycho Brahe's assistants were using it by about 1580, and it was the standard technique until Napier published logarithms in 1614.

The four formulas from the previous step start with a product and end with a sum. Most of the time you want the reverse. You are handed sin C + sin D and you want a product.

Reading the formulas right to left almost does that. The obstacle is the naming. The right hand sides are written in terms of A+B and A−B, and what you were handed is two angles with no A and B attached to them. So rename.

That is why the letters change on this page. On the compound angle page, A and B were the two angles you were adding. Here the two angles you are given are the sum and the difference of those, so they get the next two letters. A and B are the halves. C and D are the sum and the difference.

Set C = A+B and D = A−B. Add those two equations: C + D = 2A, so A = (C+D)/2. Subtract them: C − D = 2B, so B = (C−D)/2. Nothing is lost in the renaming. Any pair of angles C and D can be written this way, because the two formulas for A and B always give an answer.

Substitute into the four and you get the four everyone quotes:

sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2)

sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2)

cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2)

cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2)

These are the sum to product The four identities that rewrite a sum or difference of two sines or cosines as a product: sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2), sin C − sin D = 2 cos((C+D)/2) sin((C−D)/2), cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2), and cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2). They are the product to sum identities with A renamed to (C+D)/2 and B to (C−D)/2, so every right hand side is built from half the sum and half the difference of the two angles. Useful because a product is zero when either factor is zero, which solves equations, and because a product of waves integrates term by term. formulas. The first three substitutions are the same move three times over and mislead nobody. The fourth is the one worth watching, because the minus sign on it is not decoration and it is not a typo:

2 sin A sin B = cos(A − B) − cos(A + B) Value 0.524 = 0.524.

Substitute A = (C+D)/2 and B = (C−D)/2. Then A + B is C and A − B is D. 2 sin((C+D)/2) sin((C−D)/2) = cos D − cos C Value 0.524 = 0.524.

That is the difference the wrong way round. Multiply both sides by −1. Both sides change sign together, so the value column flips here and the equation stays true. −2 sin((C+D)/2) sin((C−D)/2) = cos C − cos D Value −0.524 = −0.524.

Read it the other way round, which is the form everyone quotes. cos C − cos D = −2 sin((C+D)/2) sin((C−D)/2) Value −0.524 = −0.524.

Check it at C = 60 degrees and D = 0 with the control above; the right hand column does the arithmetic live. Every line reads the same number on both sides, and the number changes sign at the third line, because multiplying an equation by minus 1 changes both sides at once and leaves it true.

Drop that minus sign from the formula and you would be claiming a positive number where the left side is negative, which is wrong at every angle where the product is not zero.

Now read all four in words. Every right hand side is 2, times a function of half the sum, times a function of half the difference. That structure never changes. Half the sum and half the difference, every time. The only question is which function goes where.

Two facts settle most of it. A difference must be zero when C = D, so both difference formulas carry sin((C−D)/2), which is 0 when C = D. Both sum formulas carry cos((C−D)/2) instead, which is 1 when C = D, so a sum collapses to twice the common value. And the factor on the average keeps the function you started with for a sum (sines give a sine, cosines give a cosine) and swaps it for a difference.

That pattern is worth having, but it is not worth trusting under pressure. Re-deriving the formula takes the two lines from the previous step and about ten seconds, and a formula you can re-derive is one you cannot misremember.

A sum of two waves is hard to work with. A product of two waves is easy. Two specific reasons.

First, a product is zero exactly when one of its factors is zero. That turns solving into reading. Second, a product of two sines or two cosines can be converted back into a sum of two single waves, and a single wave integrates in one step.

Take the equation sin 5x + sin 3x = 0. Guessing values gets you a few solutions and no confidence that you have them all. Factor it instead. Half the sum of the two angles is (5x + 3x)/2 = 4x. Half the difference is (5x − 3x)/2 = x. So:

sin 5x + sin 3x = 2 sin 4x cos x

The equation is now 2 sin 4x cos x = 0, and a product is zero when a factor is zero. So sin 4x = 0 or cos x = 0. Sine is zero at whole multiples of π, so 4x = kπ, giving x = kπ/4 for every integer k. Cosine is zero at π/2 plus whole multiples of π, so x = π/2 + kπ.

One honest note about this particular example. The cosine solutions are already in the sine list: π/2 is 2 × π/4, and 3π/2 is 6 × π/4. So the complete answer here is x = kπ/4, and the second factor adds nothing new. That is a coincidence of these two frequencies, not a general rule. Change the equation to sin 5x + sin 2x = 2 sin(7x/2) cos(3x/2) and the two lists have no member in common.

The other direction serves integration. ∫ sin 5x sin 3x dx has no obvious antiderivative as it stands. Use 2 sin A sin B = cos(A−B) − cos(A+B) to rewrite the integrand as (cos 2x − cos 8x)/2, and now each term is a single cosine. The integral is (sin 2x)/4 − (sin 8x)/16, plus a constant.

That calculation is the machinery underneath Fourier series. Integrate sin 5x sin 3x over one full turn and both cosine terms give zero, because each completes a whole number of cycles across that interval. Two sine waves of different whole-number frequencies multiply to something whose average is zero. That is what lets a signal be split into frequency components independently of each other, which is the result quoted on page 6.

Play two notes at almost the same pitch and you hear a single note that gets louder and quieter, over and over. The wobble is called beats The slow rise and fall in loudness heard when two tones of nearly equal frequency sound together. Adding two equal waves at frequencies f₁ and f₂ gives 2 sin(2π·(f₁+f₂)/2·t) · cos(2π·(f₁−f₂)/2·t), a tone at the average frequency multiplied by a cosine at half the difference. The beat rate is |f₁ − f₂|, not half of it, because the envelope reaches full size twice per cycle of that cosine, once at +1 and once at −1., and the sum-to-product formula is the whole explanation.

Two waves of equal size at frequencies f₁ and f₂ give the sum sin(2πf₁t) + sin(2πf₂t). Feed that into sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2), with C = 2πf₁t and D = 2πf₂t. Half the sum of the angles is 2π × ((f₁+f₂)/2) × t. Half the difference is 2π × ((f₁−f₂)/2) × t. So the sum of the two waves is:

2 sin(2π × (f₁+f₂)/2 × t) × cos(2π × (f₁−f₂)/2 × t)

Read the two factors separately. The sine is a wave at the average of the two frequencies. That is the pitch you hear. The cosine is a wave at half the difference of the two frequencies. If the two frequencies are close, half their difference is tiny, so this factor changes very slowly compared with the first.

A slowly changing factor multiplying a fast wave acts as a varying amplitude. The pair of curves +2cos(...) and −2cos(...) traces the boundary the fast wave stays inside, and that boundary is called the envelope The pair of curves that bound a wave whose amplitude is changing, traced by the slowly varying factor and its negative. For the sum of two nearly equal waves the envelope is ±2cos(2π·(f₁−f₂)/2·t). The fast wave touches the envelope at every peak and the envelope pinches to zero between beats, where the fast wave changes sign..

Now the rate, which is where people go wrong. The cosine factor has frequency (f₁−f₂)/2. The loudness does not follow the cosine, it follows the size of the cosine, and the size is largest twice per cycle: once when the cosine is at +1 and once when it is at −1. Both are loud.

So the envelope has two bulges per cycle of the cosine, and the wobbles arrive at twice the cosine's frequency. Twice (f₁−f₂)/2 is f₁−f₂. The beat rate is the difference of the two frequencies, not half of it.

Concretely, play 440 Hz against 444 Hz. You hear a single tone at 442 Hz, the average, getting louder and quieter 4 times a second. A piano tuner counts those wobbles, and 4 per second means the two strings are 4 Hz apart. Quoting 2 per second, the frequency of the cosine, is the classic error on this page.

Two limits worth stating. Between two bulges the envelope passes through zero, and at that instant the fast wave flips its sign, because the cosine has gone negative and a negative multiplier inverts the wave. The ear hears the size and so hears a gap, not a flip.

And this exact formula needs both waves to be the same size. Unequal waves still beat, but the envelope never reaches zero, so the quiet part is quiet rather than silent.

Put two unit vectors on the circle, one at angle C and one at angle D. Their components are (cos C, sin C) and (cos D, sin D), from the unit circle definition on page 2.

Add them. The x component of the sum is cos C + cos D. The y component is sin C + sin D. Both of those are left hand sides of formulas from two steps ago, so both turn into products:

cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2) 2 cos((C+D)/2) is 1.286. cos((C−D)/2) is 0.940. Value 1.208 = 1.208.

The y component, by the sine formula from the same step. The right hand factor is the same one, and it sits in the same column. sin C + sin D = 2 sin((C+D)/2) cos((C−D)/2) 2 sin((C+D)/2) is 1.532. cos((C−D)/2) is 0.940. Value 1.440 = 1.440.

The right hand column is the same on both lines. Take it out in front, and what is left is the unit vector at the average angle. (x, y) = 2 cos((C−D)/2) (cos((C+D)/2), sin((C+D)/2)) Value 1.879 = 1.879.

So the sum of the two vectors The single vector obtained by adding two or more vectors, drawn as the diagonal of the parallelogram they span. For two unit vectors at angles C and D the resultant lies along the bisector, at angle (C+D)/2, with signed length 2cos((C−D)/2), which is the sum to product identities read as the x and y components of one addition. points at the average of the two angles, and the factor taken out in front is its signed length.

The geometry says the same thing without algebra. Both vectors have length 1, so the parallelogram they span has four equal sides, which makes it a rhombus. The diagonal of a rhombus cuts its corner angle exactly in half. So the sum points along the bisector, which is the average angle.

As for the length, each vector makes an angle of half the difference with that bisector, so each contributes cos((C−D)/2) along it, and there are two of them.

One caveat about the length. cos((C−D)/2) goes negative once the two vectors are more than 180 degrees apart, and a length is not negative. Read the formula as giving a signed length along the direction (C+D)/2: when it is negative the resultant points the opposite way, at (C+D)/2 plus 180 degrees.

At exactly 180 degrees apart the half difference is 90 degrees, the cosine is 0, the two vectors cancel, and the sum is the zero vector, which has no direction at all.

This is why the average-and-half-difference pattern is worth seeing rather than memorising. The sine formula and the cosine formula are not two facts. They are the y component and the x component of one vector addition.