Sum and Difference Formulas

Summary: Continuing with trig identities, this page looks at the sum and difference formulas, namely sin(A ± B), cos(A ± B), and tan(A ± B). Remember one, and all the rest flow from it. There’s also a beautiful way to get them from Euler’s formula.

Sine and Cosine of A ± B

Formulas for cos(A + B), sin(A − B), and so on are important but hard to remember. Yes, you can derive them by strictly trigonometric means. But such proofs are lengthy, too hard to reproduce when you’re in the middle of an exam or of some long calculation.

This brings us to W.W. Sawyer’s marvelous idea, as expressed in chapter 15 of Mathematician’s Delight (1943; reprinted 1991 by Penguin Books). He shows how you can derive the sum and difference formulas by ordinary algebra and one simple formula.

The ordinary algebra is simply the rules for combining powers:

(If you’re a bit rusty on the laws of exponents, you may want to review them.)

Euler’s Formula

You may already know the “simple formula” that I mentioned above. It’s

The formula is not Sawyer’s, by the way; it’s commonly called Euler’s formula. I don’t even know whether the idea of using Euler’s formula to get the sine and cosine of sum and difference is original with Sawyer. But I’m going to give him credit, since his explanation is simple and clear and I’ve never seen it explained in this way anywhere else.

You’ll sometimes see cos x + i sin x abbreviated as cis x for brevity.

I’ve marked Euler’s formula “memorize”. Although it’s not hard to derive (and Sawyer does it in a few steps by means of power series), you have to start somewhere. And that formula has so many other applications that it’s well worth committing to memory. For instance, you can use it to get the roots of a complex number and the logarithm of a negative number.

Sine and Cosine of a Sum

Okay, back to Sawyer’s idea. What happens if you substitute x = A + B in equation 47 above? You get

cos(A + B) + i sin(A + B) = e^{iA + iB}

Hmmm, this looks interesting. It involves exactly what we’re looking for, cos(A + B) and sin(A + B). Can you simplify the right-hand side? Yes, use equation 46 and then equation 47 to rewrite it:

cos(A + B) + i sin(A + B) = e^{iA + iB}

cos(A + B) + i sin(A + B) = e^iA e^iB

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

Multiply out the right-hand side, and group real and imaginary terms separately:

cos(A + B) + i sin(A + B) = cos A cos B + i sin A cos B + i cos A sin B + i² sin A sin B

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

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

Now here’s the sneaky part. If I tell you a+bi = 7−9i and ask you to solve for a and b, you know immediately that a = 7 and b = −9, right? More formally, if two complex numbers are equal, their real parts must be equal and their imaginary parts must be equal. So the above equation in sines and cosines is actually two equations, one for the real part and one for the imaginary part. (I’m showing the imaginary part first in the box below, to put sine before cosine.)

In just a few short steps, the formulas for cos(A + B) and sin(A + B) flow right from equation 47, Euler’s equation for e^i x. No more need to memorize which one has the minus sign and how all the sines and cosines fit on the right-hand side: all you have to do is a couple of substitutions and a multiply.

Example: What’s the exact value of cos 75° or cos(5π/12)?

Solution: 75° = 45°+30° (5π/12 = π/4+π/6). Using equation 48,

cos 75° = cos(45°+30°)

cos 75° = cos 45° cos 30° − sin 45° sin 30°

cos 75° = (√2/2)×(√3/2) − (√2/2)×(1/2)

cos 75° = √6/4 − √2/4

Sine and Cosine of a Difference

What about the formulas for the differences of angles? You can write them down at once from equation 48 by substituting −B for B and using equation 22. Or, if you prefer, you can get them by substituting x = A−B in equation 47 above. Either way, you get

Some Geometric Proofs

I personally find the algebraic reasoning given above very easy to follow, though you do have to remember Euler’s formula.

If you prefer geometric derivations of sin(A ± B) and cos(A ± B), you’ll find a beautiful set by Len and Deborah Smiley. (Phil Kenny drew my attention to this page’s original version and to the link at the University of Alaska.) Eric’s Treasure Trove of Mathematics has smaller versions of the pictures.

The fallback position is the standard proof: draw a diagram and use the distance formula or Pythagorean Theorem to prove the formula for cos(A − B).

Tangent of A ± B

Sometimes—though not very often—you have to deal with the tangent of the sum or difference of two angles. I have only a vague idea of the formula, but it’s easy enough to work out “on the fly”:

tan(A + B) = sin(A + B) / cos(A + B)

tan(A + B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)

What a mess! There’s no way to factor that and remove common terms—or is there? Suppose you start with a vague idea that you’d like to know tan(A+B) in terms of tan A and tan B rather than all those sines and cosines. The numerator and denominator contain sines and cosines, so if you divide by cosines you’d expect to end up with sines or perhaps sines over cosines. And sine/cosine is tangent, so this seems like a promising line of attack. Since you’ve got cosines of angles A and B to contend with, try dividing the numerator and denominator of the fraction by cos A cos B:

tan(A + B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)

tan(A + B) = [sin A/cos A + sin B/cos B] / [1 − (sin A/cos A)(sin B/cos B)]

Success! Simplify it using the definition of tan x, and you have

Now if you replace B with −B, you have the formula for tan(A − B). (Take a minute to review why tan(−x) = −tan x.)

Example: What’s the exact value of tan 15° or tan(π/12)?

Solution: 15° = 60°−45° (π/12 = π/3 − π/4). Therefore

tan(π/12) = tan(π/3 − π/4)

tan(π/12) = [tan(π/3) − tan(π/4)] / [1 + tan(π/3) tan(π/4)]

tan(π/12) = (√3− 1) / (1 + √3×1)

tan(π/12) = (√3− 1) / (√3 + 1)

If you like, you can rationalize the denominator:

tan(π/12) = (√3− 1)² / (√3 + 1)(√3 − 1)

tan(π/12) = (3 − 2√3 + 1) / (3 − 1)

tan(π/12) = (4 − 2√3) / 2

tan(π/12) = 2 − √3

Practice Problems

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

BTW: Product-Sum Formulas

Product to Sum

Sometimes you need to simplify an expression like cos 3x cos 5x. Of course it’s not equal to cos(15x²), but can it be simplified at all? The answer is yes, and in fact you need this technique for calculus work. There are four formulas that can be used to break up a product of sines or cosines.

These product-to-sum formulas come from equation 48 and equation 49 for sine and cosine of A ± B. First let’s develop one of these formulas, and then we’ll look at an application before developing the others.

Take the two formulas for cos(A ± B) and add them:

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

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

cos(A − B) + cos(A + B) = 2 cos A cos B

½ [cos(A − B) + cos(A + B)] = cos A cos B

Example: Suppose you need to graph the function

f(x) = cos 2x cos 3x,

or perhaps you need to find its integral. Both of these are rather hard to do with the function as it stands. But you can use the product-to-sum formula, with A = 2x and B = 3x, to rewrite the function as a sum:

f(x) = cos 2x cos 3x

f(x) = ½ [cos(2x − 3x) + cos(2x + 3x)]

f(x) = ½ [cos(−x) + cos 5x]

You know that cos(−x) = cos x, and therefore

f(x) = ½ [cos x + cos 5x]

f(x) = ½ cos x + ½ cos 5x

This is quite easy to integrate. And while it’s not exactly trivial to graph, it’s much easier than the original, because cos x and cos 5x are easy to graph.

The other three product-to-sum formulas come from the other three ways to add or subtract the formulas in equation 48 and equation 49. If you subtract the two cosine formulas instead of adding:

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

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

you get

cos(A − B) − cos(A + B) = 2 sin A sin B

½ [cos(A−B)  −  cos(A + B)] = sin A sin B

To get the other two product-to sum formulas, add the two sine formulas from equation 48 and equation 49, or subtract them. Here are all four formulas together:

The fourth one of those formulas really isn’t needed, because you can always evaluate cos p sin q as sin q cos p. But it’s traditional to present all four formulas.

Sum to Product

There are also formulas that combine a sum or difference into a product. Heon Joon Choi, a physics student from Cornell, has kindly told me of an application: “superposing two waves and trying to figure out the nodes is much easier if they are multiplied, rather than added.” This makes sense: solving most equations is easier once you’ve factored them. The sum-to-product formulas are also used to prove the Law of Tangents, though that itself is no longer used in solving triangles.

Here’s how to get the sum-to-product formulas. First make these definitions:

A = ½(u + v), and B = ½(u − v)

Then you can see that

A + B = u, and A − B = v

Now make those substitutions in all four formulas of equation 52, and after simplifying you will have the sum-to-product formulas:

BTW: Proof of Euler’s Formula

Euler’s formula (equation 47) is easily proved by means of power series. Start with the formulas

You can click the image to see a larger version. Or if you can’t see the image at all, here are the formulas in linear text:

e^x = SUM [ x^k / k! ] = 1 + x + x²/2! + x³/3! + ⋯

cos x = SUM [ (−1)^k x^2k / (2k)! ] = 1 − x²/2! + x⁴/4! − x⁶/6! + ⋯

sin x = SUM [ (−1)^k x^2k+1 / (2k+1)! ] = x − x³/3! + x⁵/5! − x⁷/7! + ⋯

(These are how the function values are actually calculated, by the way. All three series converge quickly, meaning that you get quite an accurate result from computing just the first few terms. If you want to know the value of e², you just substitute 2 for x in the formula and compute until the additional terms fall within your desired accuracy.)

Now we have to find the value of e^ix, where i = √−1. Use the first formula to find e^ix, by substituting ix for x:

e^ix = SUM [ (ix)^k / k! ]

e^ix = 1 + (ix) + (ix)²/2! + (ix)³/3! + (ix)⁴/4! + (ix)⁵/5! + (ix)⁶/6! + (ix)⁷/7! + ...

Simplify the powers of i, using i² = −1:

e^ix = 1 + ix − x²/2! − ix³/3! + x⁴/4! + ix⁵/5! − x⁶/6! − ix⁷/7! + ...

Finally, group the real and imaginary terms separately:

e^ix = [1 − x²/2! + x⁴/4! − x⁶/6! + ...] + i[x − x³/3! + x⁵/5! − x⁷/7! + ...]

Those should look familiar. If you refer back to the power series at the start of this section, you’ll see that the first group of terms is just cos x and the second group is just sin x. So you have

e^ix = cos x + i sin x

which is Euler’s formula, as advertised!

You may wonder where the series for cos x, sin x, and e^x come from. The answer is that they are the Taylor series expansions of the functions. (You’ll probably study Taylor series in second- or third-semester calculus.)

next:  8/Double and Half Angles