Product-to-Sum and Sum-to-Product

When Two Waves Meet

Pluck two guitar strings tuned almost the same and you hear a slow “wah… wah… wah” pulsing over the note. Those pulses are called beats, and they are pure trigonometry.

The top two rows are the separate waves, each a steady sine. Add them and you get the bottom wave (green), which speeds along inside a slow envelope (yellow) and pulses loud, soft, loud, soft. Adding two sines produced a product: a fast wave times a slow one. That is the whole idea of this lesson.


Sums into Products

When you add two sines or two cosines, you can rewrite the sum as a product:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}

The first factor uses the average of the two angles. The second uses half their difference.

In the beats animation, A+B2\frac{A+B}{2} is the fast wave you hear as the tone, and cos⁡A−B2\cos\frac{A-B}{2} is the slow envelope you hear as the pulsing. When AA and BB are close, that difference is tiny, so the envelope is very slow. That is why slightly-out-of-tune strings beat slowly.


The Differences Too

Subtraction works the same way, it just swaps a sine and a cosine:

sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}

cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

Watch the minus sign in front of the cosine difference. It is easy to drop.


Products into Sums

The reverse direction is just as useful. A product of sines and cosines becomes a sum:

sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A \cos B = \tfrac{1}{2}\big[\sin(A+B) + \sin(A-B)\big]

cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]\cos A \cos B = \tfrac{1}{2}\big[\cos(A-B) + \cos(A+B)\big]

sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A \sin B = \tfrac{1}{2}\big[\cos(A-B) - \cos(A+B)\big]


Where They Come From

Every one of these is just the addition and subtraction formulas, added or subtracted.

Line up the two sine formulas:

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\begin{aligned} \sin(A+B) &= \sin A \cos B + \cos A \sin B \\[0.5em] \sin(A-B) &= \sin A \cos B - \cos A \sin B \end{aligned}

Add them. The cos⁡Asin⁡B\cos A \sin B terms cancel:

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

Divide by 2 and you have the product-to-sum rule for sin⁡Acos⁡B\sin A \cos B. The sum-to-product rules are the same identities read backwards, letting A+BA + B and A−BA - B become the new angles.


Example: Product to Sum

Rewrite 2sin⁡75°cos⁡15°2\sin 75° \cos 15° as a sum.

2sin⁡75°cos⁡15°=sin⁡(75°+15°)+sin⁡(75°−15°)=sin⁡90°+sin⁡60°=1+32\begin{aligned} 2\sin 75° \cos 15° &= \sin(75° + 15°) + \sin(75° - 15°) \\[0.5em] &= \sin 90° + \sin 60° \\[0.5em] &= 1 + \frac{\sqrt{3}}{2} \end{aligned}

A messy product turned into two clean, known values.


The Full Set

Sum to product:

SumProduct
sin⁡A+sin⁡B\sin A + \sin B2sin⁡A+B2cos⁡A−B22\sin\frac{A+B}{2}\cos\frac{A-B}{2}
sin⁡A−sin⁡B\sin A - \sin B2cos⁡A+B2sin⁡A−B22\cos\frac{A+B}{2}\sin\frac{A-B}{2}
cos⁡A+cos⁡B\cos A + \cos B2cos⁡A+B2cos⁡A−B22\cos\frac{A+B}{2}\cos\frac{A-B}{2}
cos⁡A−cos⁡B\cos A - \cos B−2sin⁡A+B2sin⁡A−B2-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

Product to sum:

ProductSum
sin⁡Acos⁡B\sin A \cos B12[sin⁡(A+B)+sin⁡(A−B)]\tfrac{1}{2}[\sin(A+B) + \sin(A-B)]
cos⁡Acos⁡B\cos A \cos B12[cos⁡(A−B)+cos⁡(A+B)]\tfrac{1}{2}[\cos(A-B) + \cos(A+B)]
sin⁡Asin⁡B\sin A \sin B12[cos⁡(A−B)−cos⁡(A+B)]\tfrac{1}{2}[\cos(A-B) - \cos(A+B)]