Notes and Digressions

The Problem with Memorizing

Dad sighed. “Kip, do you think that table was brought down from on high by an archangel?”

Robert A. Heinlein, in Have Space Suit—Will Travel (1958)

It’s not just that there are so many trig identities; they seem so arbitrary. Of course they’re not really arbitrary, since all can be proved; but when you try to memorize all of them they seem like a jumble of symbols where the right ones aren’t more obviously right than the wrong ones. For example, is it sec²A = 1 + tan²A or tan²A = 1 + sec²A ? I doubt you know off hand which is right; I certainly don’t remember. Who can remember a dozen or more like that, and remember all of them accurately?

Too many teachers expect (or allow) students to memorize the trig identities and parrot them on demand, much like a series of Bible verses. In other words, even if they’re originally taught as a series of connected propositions, they’re remembered and used as a set of unrelated facts. And that, I think, is the problem. The trig identities were not brought down by an archangel; they were developed by mathematicians, and it’s well within your grasp to re-develop them when you need to. With effort, we can remember a few key facts about anything. But it’s much easier if we can fit them into a context, so that the identities work together as a whole.

Why bother? Well, of course it will make your life easier in trig class. But you’ll also need the trig identities in later math classes, especially calculus, and in physics and engineering classes. In all of those, you’ll find the going much easier if you’re thoroughly grounded in trigonometry as a unified field of knowledge instead of a collection of unrelated facts.

This is why it’s easier to remember almost any song than an equivalent length of prose: the song gives you additional cues in the rhythm, common patterns of emphasis, and usually rhymes at the ends of lines. With prose you have only the general thought to hold it together, so that you must memorize it essentially as a series of words. With the song there are internal structures that help you, even if you’re not aware of them.

If you’re memorizing Lincoln’s Gettysburg Address, you might have trouble remembering whether he said “recall” or “remember” at a certain point; in a song, there’s no possible doubt which of those words is right because the wrong one won’t fit in the rhythm.

On the Other Hand ...

I’m not against all memorization. Some things have to be memorized because they’re a matter of definition. Others you may choose to memorize because you use them very often, you’re confident you can memorize them correctly, and the derivation takes more time than you’re comfortable with. Still others you may not set out to memorize, but after using them many times you find you’ve memorized them without trying to—much like a telephone number that you dial often.

I’m not against all memorization; I’m against needless memorization used as a substitute for thought. If you decide in particular cases that memory works well for you, I won’t argue. But I do hope you see the need to be able to re-derive things on the spot, in case your memory fails. Have you ever dialed a friend’s telephone number and found you couldn’t quite remember whether it was 6821 or 8621? If you can’t remember a phone number, you have to look it up in the book. My goal is to free you from having to look up trig identities in the book.

Thanks to David Dixon at [email protected] for an illuminating exchange of notes on this topic. He made me realize that I was sounding more anti-memory than I meant to, and in consequence I’ve added this note. But he may not necessarily agree with what I say here.

I wrote these pages to show you how to make the trig identities “fit” as a coherent whole, so that you’ll have no more doubt about them than you do about the words of a song you know well. The difference is that you won’t need to do it from memory. And you’ll gain the sense of power that comes from mastering your subject instead of groping tentatively and hoping to strike the right answer by good luck.

Proof of Euler’s Formula

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

(These are how the function values are actually calculated, by the way. 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)ⁿ / n! ]

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.)

Polar Form of a Complex Number

As you may remember, complex numbers like 3+4i and 2−7i are often plotted on a complex plane. This makes it easier to visualize adding and subtracting. The illustration at right shows that

(3+4i) + (2−7i) = 5−3i

It turns out that for multiplying, dividing, and finding powers and roots, complex numbers are easier to work with in polar form. This means that instead of thinking about the real and imaginary components (the “−3” and “5” in −3+5i), you think of the length and direction of the line.

The length is easy, just the good old Pythagorean theorem in fact:

r = √(3²+5²) = √34 ≈ 5.83

(The length r is called the absolute value or modulus of the complex number.)

But what about the direction? You can see from the picture that θ about 120° or about 2 radians (measured counterclockwise from the positive real axis, which points east), but what is it exactly? Well, you know that tan θ = −5/3. If you take Arctan(−5/3) on your calculator, you get −1.03, and adding π to get into the proper quadrant gives θ = 2.11 radians. (In degrees, Arctan(−5/3) = −59.04, and adding 180 gives θ = 120.96°.) You can write it this way:

−3+5i ≈ 5.83 e^2.11i or 5.83[cos 2.11 + i sin 2.11] or 5.83 cis 2.11 or 5.83∠120.96°

There’s a nice trick that finds the angle in the correct quadrant automatically:

θ = 2 Arctan(y/(x+r))

This builds in the adjustment to the correct quadrant, so you never have to worry about whether to add or subtract 180° or π. If you test 2 Arctan(5/(−3+√34)) on your calculator, you get 2.11 radians or 120.96° as before.

This formula is particularly handy when you’re writing a program or spreadsheet formula to find θ. It flows directly from the formulas for tangent of half an angle, where the functions are expressed in terms of x, y, and r. A July 2003 article by Rob Johnson, archived here, inspired me to add this section.

Powers and Roots of a Complex Number

One of the applications of Euler’s formula (equation 47) is finding any power or root of any complex number. (Sawyer doesn’t do this, or at least not in the same book.) It’s not hard to develop the formula, using just Euler’s formula and the laws of exponents.

Start by writing down Euler’s formula, multiplied left and right by a scale factor r:

r (cos x + i sin x) = r e^ix

Next raise both sides to the Nth power:

[r (cos x + i sin x)]^N = [r e^ix]^N

Let’s leave the left-hand side alone for a while, and simplify the right-hand side. First, by the laws of exponents

[r e^ix]^N = r^N e^iNx

Apply Euler’s formula to the right-hand side and you have

[r e^ix]^N = r^N (cos Nx + i sin Nx)

Connect that right-hand side to the original left-hand side and you have DeMoivre’s theorem:

This tells you that if you put a number into polar form, you can find any power or root of it. (Remember that the nth root of a number is the same as the 1/n power.)

Square Root of i

For instance, you know the square roots of −1 are i and −i, but what’s the square root of i? You can use DeMoivre’s theorem to find it.

First, put i into e^ix form (polar form), using the above technique. i = 0+1i. cos x = 0 and sin x = 1 when x = 90° or π/2. Therefore

i = cos(π/2) + i sin(π/2)

√i = i^½ = [cos(π/2) + i sin(π/2)]^½

And by equation 82

√i = cos(½×π/2) + i sin(½×π/2)

√i = cos(π/4) + i sin(π/4)

√i = 1/√2 + i×1/√2

√i = (1+i)/√2

The other square root is minus that, as usual.

Principal Root of Any Number

You can find any root of any complex number in a similar way, but usually with one preliminary step.

For instance, suppose you want the cube roots of 3+4i. The first step is to put that number into polar form. The absolute value is sqrt(3²+4²) = 5, and you use the Arctan technique given above to find the angle θ = 2 Arctan(4/(3+5)) = 2 Arctan(½) ≈ 53.13° or 0.9273 radians. The polar form is

3+4i ≈ 5(cos 53.13° + i sin 53.13°)

To take a cube root of that, use equation 82 with N = 1/3:

cube root of (3+4i) = 5^1/3 [cos(53.13°/3) + i sin(53.13°/3)]

The cube root of 5 is about 1.71. Dividing the angle θ by 3, 53.13°/3 ≈ 17.71°; the cosine and sine of that are about 0.95 and 0.30. Therefore

cube root of (3+4i) ≈ 1.71 × (cos 17.71° + i sin 17.71°)

cube root of (3+4i) ≈ 1.63 + 0.52i

Multiple Roots

You may have noticed that I talked about “the cube roots [plural] of 3+4i” and what we found was “a cube root”. Even with the square root of i, I waved my hand and said that the “other” square root was minus the first one, “as usual”.

You already know that every positive real has two square roots. In fact, every complex number has n nth roots.

How can you find them? It depends on the periodic nature of the cosine and sine functions. To start, look back at Euler’s formula,

e^ix = cos x + i sin x

What happens if you add 2π or 360° to x? You have

e^i(x+2π) = cos(x+2π) + i sin(x+2π)

But taking sine or cosine of 2π plus an angle is exactly the same as taking sine or cosine of the original angle. So the right-hand side is equal to cos x + i sin x, which is equal to e^ix. Therefore

e^i(x+2π) = e^ix

In fact, you can keep adding 2π or 360° to x as long as you like, and never change the value of the result. Symbolically,

e^i(x+2πk) = e^ix   for all integer k

When you take an nth root, you simply use that identity.

Getting back to the cube roots of 3+4i, recall that that number is the same as 5e^iθ, where θ is about 53.13° or 0.9273 radians. The three cube roots of e^iθ are

e^iθ/3, e^i(θ+2π)/3, and e^i(θ+4π)/3   for k = 0,1,2

or

e^iθ/3, e^{i(θ/3+2π/3)}, and e^{i(θ/3+4π/3)}   for k = 0,1,2

Compute those as cos x + i sin x in the usual way, and then multiply by the (principal) cube root of 5. I get these three roots:

cube roots of 3+4i ≈ 1.63+0.52i, −1.26+1.15i, −0.36−1.67i

More generally, we can extend DeMoivre’s theorem (equation 82) to show the N Nth roots of any complex number:

Logarithm of a Negative Number

Another application flows from a famous special case of Euler’s formula. Substitute x = π or 180° in equation 47. Since sin 180° = 0, the imaginary term drops out. And cos 180° = −1, so you have the famous formula

It’s interesting to take the natural log of both sides:

ln(−1) = ln(e^iπ)

which gives

ln(−1) = iπ

It’s easy enough to find the logarithm of any other negative number. Since

ln(ab) = ln a + ln b

then for all real a you have

ln(−a) = ln[a × −1] = ln a + ln(−1) = ln a + iπ

I don’t honestly know whether all of this has any practical application. But if you’ve ever wondered about the logarithm of a negative number, now you know.

Cool Proof of Double-Angle Formulas

I can’t resist pointing out another cool thing about Sawyer’s marvelous idea: you can also use it to prove the double-angle formulas equation 59 directly. From Euler’s formula for e^ix you can immediately obtain the formulas for cos(2A) and sin(2A) without going through the formulas for sums of angles. Here’s how.

Remember the laws of exponents: x^ab = (x^a)^b. One important special case is that x^2b = (x^b)². Use that with Euler’s formula (equation 47):

cos(2A) + i sin(2A) = e^(2A)i

   = (e^iA)²

   = (cos A + i sin A)²

   = cos²A + 2i sinA cosA + i²sin²A

   = cos²A + 2i sinA cosA − sin²A

   = cos²A − sin²A + 2i sinA cosA

Since the real parts on left and right must be equal, you have the formula for cos(2A). Since the imaginary parts must be equal, you have the formula for sin(2A). That’s all there is to it.

Multiple-Angle Formulas

The above technique is even more powerful for deriving formulas for functions of 3A, 4A, or any multiple of angle A. To derive the formulas for nA, expand the nth power of (cos A + i sin A), then collect real and imaginary terms.

This was inspired by a May 2009 reading of Paul Nahin’s An Imaginary Tale: The Story of √−1 (Princeton, 1998).

Just to show the method, I’ll derive the functions of 3A and 4A. You can try the brute-force approach of cos(2A+2A) and sin(2A+2A) and see how much effort my method saves.

Functions of 3A

De Moivre’s theorem (equation 82) tells us that

cos 3A + i sin 3A = (cos A + i sin A)³

Expand the right-hand side via the binomial theorem, remembering that i² = −1 and i³ = −i:

cos 3A + i sin 3A = cos³A + 3 i cos²A sin A − 3 cos A sin²A − i sin³A

Collect real and imaginary terms:

cos 3A + i sin 3A = cos³A − 3 cos A sin²A + 3 i cos²A sin A − i sin³A

cos 3A + i sin 3A = [cos A (cos²A − 3 sin²A)] + i [sin A (3 cos²A − sin²A)]

Set the real part on the left equal to the real part on the right, and similarly for the imaginary parts, and you have the formulas for the cosine and sine of 3A:

cos 3A = cos A (cos²A − 3 sin²A) and sin 3A = sin A (3 cos²A − sin²A)

The tangent formula is easy to get: just divide.

tan 3A = sin 3A / cos 3A

tan 3A = [sin A (3 cos²A − sin²A)] / [cos A (cos²A − 3 sin²A)]

tan 3A = tan A (3 cos²A − sin²A) / (cos²A − 3 sin²A)

Divide top and bottom by cos²A to simplify:

tan 3A = tan A (3 − tan²A) / (1 − 3 tan²A)

Functions of 4A

It’s no more work to find the functions of 4A. I’ll just run through the steps without commentary. First, cosine and sine of 4A:

cos 4A + i sin 4A = (cos A + i sin A)⁴

cos 4A + i sin 4A = cos⁴A + 4 i cos³A sin A − 6 cos²A sin²A − 4 i cos A sin³A + sin⁴A

cos 4A + i sin 4A = cos⁴A − 6 cos²A sin²A + sin⁴A + 4 i cos³A sin A − 4 i cos A sin³A

cos 4A = cos⁴A − 6 cos²A sin²A + sin⁴A and sin 4A = 4 cos³A sin A − 4 cos A sin³A

cos 4A = (cos²A − sin²A)² − 4 sin²A cos²A and sin 4A = 4 sin A cos A (cos²A − sin²A)

Now the tangent formula:

tan 4A = sin 4A / cos 4A

tan 4A = [4 sin A cos A (cos²A − sin²A)] / [(cos²A − sin²A)² − 4 sin²A cos²A]

Divide top and bottom by cos⁴A:

tan 4A = 4 tan A (1 − tan²A) / [(1 − tan²A)² − 4 tan²A]

Great Book on Problem Solving

I have to recommend a terrific little book, How To Solve It by G. Polya. Most teachers aren’t very good at teaching you how to solve problems and do proofs. They show you how they do them, and expect you to pick up their techniques by a sort of osmosis. But most of them aren’t very good at explaining the thought process that goes into doing a geometrical proof, or solving a dreaded “story problem”.

Polya’s book does a great job of teaching you how to solve problems. He shows you the kinds of questions you should ask yourself when you see a problem. In other words, he teaches you how to get yourself over the hum, past the floundering that most people do when they see an unfamiliar problem. And he does it with lots of examples, so that you can develop confidence in your techniques and compare your methods with his. The techniques I’ve mentioned above are just three out of the many in his book.

There’s even a handy checklist of questions you can ask yourself whenever you’re stuck on a problem.

How To Solve It was first published in 1945, and it’s periodically in and out of print. If you can’t get it from your bookstore, go to the library and borrow a copy. You won’t be sorry.

What’s New

• 10 Aug 2014: Add Bob Allison’s U-shaped approach to remembering the six functions.
• 13 Apr 2014: When I moved the discussion of functions of negative angles from Chapter 2 to Chapter 4, I missed part of it. Bob Allison kindly pointed this out, and it’s now moved after the equation that it refers to. I also took this opportunity to spell out the proof that tan and cot are cofunctions.
• 25 Jan 2012: Correct a substitution in the first attempt at a derivation of the Law of Cosines, thanks to Mark Schoonover.
• 5 Jul 2010: Add a disclaimer about complex trigonometry, following a note from Altan Kartaltepe.
• 19 May 2010: Add link to a Youtube video mnemonic for the basic trig identities.
• 25 Aug 2009: Correct typo in Reference Angles: 0 to 90° was listed as 0 to π/4, and that should have been 0 to π/2.
• 25 May 2009: Add Multiple-Angle Formulas.
• (intervening changes suppressed)
• 19 Feb 1997: new document

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