Trig without Tears:

# Notes and Digressions

revised 10 Aug 2014

Copyright © 1997-2014 Stan Brown, Oak Road Systems

Trig without Tears:

revised 10 Aug 2014

Copyright © 1997-2014 Stan Brown, Oak Road Systems

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
*sec*²A = 1 + *tan*²A
or *tan*²A = 1 + *sec*²A ?
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.

I’m not against all memorization.
**Some things have to be memorized** because they’re a
matter of definition. Others you may

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.

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

(81) *e*^{x} = SUM [ *x*^{n} / *n*! ] = 1 + *x* + *x*^{2}/2! + *x*^{3}/3! + ...

*cos* *x* = SUM [ (−1)^{n} *x*^{2n} / (2*n*)! ] = *1* − *x*^{2}/2! + *x*^{4}/4! − *x*^{6}/6! + ...

*sin* *x* = SUM [ (−1)^{n} *x*^{2n+1} / (2*n*+1)! ] = *x* − *x*^{3}/3! + *x*^{5}/5! − *x*^{7}/7! + ...

(These are how the function values are
actually calculated, by the way. If you want to know the value of
*e*^{2}, 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 *i*² = −1:

*e*^{ix} =
1 + *ix* −
*x*^{2}/2! − *ix*^{3}/3! +
*x*^{4}/4! + *ix*^{5}/5! −
*x*^{6}/6! − *ix*^{7}/7! + ...

Finally, group the real and imaginary terms separately:

*e*^{ix} =
[1 − *x*^{2}/2! +
*x*^{4}/4! − *x*^{6}/6! + ...] +
*i*[*x* − *x*^{3}/3! +
*x*^{5}/5! − *x*^{7}/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.)

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

(3+4*i*) + (2−7*i*) = 5−3*i*

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+5*i*), 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+5*i* ≈
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.

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* N*x* + *i* *sin* N*x*)

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

(82) [*r*(*cos* *x* + *i* *sin* *x*)]^{N} = *r*^{N} (*cos* N*x* + *i* *sin* N*x*)

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

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+1*i*.
*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.

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+4*i*. 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+4*i* ≈
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+4*i*) =
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+4*i*) ≈
1.71 × (*cos* 17.71° + *i* *sin* 17.71°)

cube root of (3+4*i*) ≈
1.63 + 0.52*i*

You may have noticed that I talked about “**the** cube
root**s** [plural] of 3+4*i*” 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 *n*th root, you simply use that
identity.

Getting back to the cube roots of 3+4*i*, 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+4*i* ≈
1.63+0.52*i*, −1.26+1.15*i*, −0.36−1.67*i*

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

(83) [*r* (*cos* *x* + *i* *sin* *x*]^{(1/N)} = *r*^{(1/N)} [*cos*(*x*/N+2πk/N) + *i* *sin*(*x*/N+2πk/N)] for k = 0,1,2,...,N−1

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

(84) −1 = *e*^{iπ}

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

*ln*(−1) = *ln*(*e*^{iπ})

which gives

*ln*(−1) = *i*π

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* + *i*π

(85) *ln*(−*a*) = *ln* *a* + *i*π

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.

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

= *cos*²A + 2*i* *sin*A *cos*A + *i*²*sin*²A

= *cos*²A + 2*i* *sin*A *cos*A − *sin*²A

= *cos*²A − *sin*²A + 2*i* *sin*A *cos*A

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.

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 *n*A, expand the
*n*th 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.

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 *i*² = −1 and
*i*³ = −*i*:

*cos* 3A + *i* *sin* 3A =
*cos*³A + 3 *i* *cos*²A *sin* A
− 3 *cos* A *sin*²A − *i* *sin*³A

Collect real and imaginary terms:

*cos* 3A + *i* *sin* 3A =
*cos*³A − 3 *cos* A *sin*²A
+ 3 *i* *cos*²A *sin* A − *i* *sin*³A

*cos* 3A + *i* *sin* 3A =
[*cos* A (*cos*²A − 3 *sin*²A)]
+ *i* [*sin* A (3 *cos*²A − *sin*²A)]

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 (*cos*²A − 3 *sin*²A)
and
*sin* 3A = *sin* A (3 *cos*²A − *sin*²A)

The tangent formula is easy to get: just divide.

*tan* 3A = *sin* 3A / *cos* 3A

*tan* 3A =
[*sin* A (3 *cos*²A − *sin*²A)] /
[*cos* A (*cos*²A − 3 *sin*²A)]

*tan* 3A =
*tan* A (3 *cos*²A − *sin*²A) /
(*cos*²A − 3 *sin*²A)

Divide top and bottom by *cos*²A to simplify:

*tan* 3A =
*tan* A (3 − *tan*²A) /
(1 − 3 *tan*²A)

(86) *cos* 3A = *cos* A (*cos*²A − 3 *sin*²A)

*sin* 3A = *sin* A (3 *cos*²A − *sin*²A)

*tan* 3A = *tan* A (3 − *tan*²A) / (1 − 3 *tan*²A)

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 =
*cos*^{4}A
+ 4 *i* *cos*³A *sin* A
− 6 *cos*²A *sin*²A
− 4 *i* *cos* A *sin*³A
+ *sin*^{4}A

*cos* 4A + *i* *sin* 4A =
*cos*^{4}A
− 6 *cos*²A *sin*²A
+ *sin*^{4}A
+ 4 *i* *cos*³A *sin* A
− 4 *i* *cos* A *sin*³A

*cos* 4A =
*cos*^{4}A
− 6 *cos*²A *sin*²A
+ *sin*^{4}A
and
*sin* 4A =
4 *cos*³A *sin* A
− 4 *cos* A *sin*³A

*cos* 4A = (*cos*²A − *sin*²A)²
− 4 *sin*²A *cos*²A and
*sin* 4A =
4 *sin* A *cos* A (*cos*²A
− *sin*²A)

Now the tangent formula:

*tan* 4A = *sin* 4A / *cos* 4A

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

Divide top and bottom by *cos*^{4}A:

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

(87) *cos* 4A = (*cos*²A − *sin*²A)² − 4 *sin*²A *cos*²A

*sin* 4A = 4 *sin* A *cos* A (*cos*²A − *sin*²A)

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

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.

**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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