Trig without Tears Part 2:

# The Six Functions

Copyright © 1997–2020 Stan Brown, BrownMath.com

**Summary:**
Every one of the six trig functions is just
**one side of a right triangle divided by another** side.
Or if you draw the triangle in a **unit circle**, every function
is the length of one line segment. The **easy way to remember** all
six definitions: **memorize the definitions of sine and cosine**
and then **remember the other four as combinations of sine and cosine**,
not as independent functions.

A picture is worth a thousand words (which is why it takes a thousand
times as long to download). The trig functions are nothing more than
**lengths of various sides of a right triangle in various ratios**.
Since there are three sides, there are 3 × 2 = 6
different ways to make a ratio (fraction) of sides. That’s
why there are **six trig functions, no more and no less**.

Of those six functions, three—sine, cosine, and
tangent—get the lion’s share of the work.
(The others are studied because they can be used to
make some expressions simpler.)
We’ll begin with sine and cosine, because they are **really basic**
and the others depend on them.

Here is one of the conventional ways of showing
a right triangle. A **key point** is that the lower-case letters
*a*, *b*, *c*
are the sides opposite to the angles marked with the corresponding
capital letters *A*, *B*, *C*. Most books use this convention:
**lower-case letter for side opposite upper-case angle**.

The two fundamental definitions are marked in the diagram.
**You must commit them to memory**.
In fact, they should become second nature to
you, so that you recognize them no matter how the triangle is turned
around. Always, always, the sine of an angle is equal to the opposite
side divided by the hypotenuse (opp/hyp in the diagram). The cosine is
equal to the adjacent side divided by the hypotenuse (adj/hyp).

(1) **Memorize:**

sine = (opposite side) / hypotenuse

cosine = (adjacent side) / hypotenuse

What is the sine of *B* in the diagram? Remember opp/hyp: the opposite
side is *b* and the hypotenuse is *c*, so
sin *B* = *b*/*c*.
What about the cosine of *B*? Remember adj/hyp: the adjacent side is
*a*, so cos *B* = *a*/*c*.

Do you notice that the sine of one angle is the cosine of the other?
Since *A* + *B* + *C* = 180° for any
triangle, and *C* is 90° in this triangle,
*A* + *B* must
equal 90°. Therefore
*A* = 90° − *B*, and
*B* = 90° − *A*. When two angles add to
90°, each angle is the **co**mplement of the other, and the sine
of each angle is the **co**sine of the other.
These are the **cofunction identities**:

(2) sin *A* = cos(90° − *A*) or cos(π/2 − *A*)

cos *A* = sin(90° − *A*) or sin(π/2 − *A*)

The definitions of sine and cosine can be rearranged a little bit to
let you write down the lengths of the sides in terms of the
hypotenuse and the angles. For example, when you know
that *b*/*c* = cos *A*, you
can multiply through by *c*
and get *b* = *c* × cos *A*. Can you write another
expression for length *b*, one that uses a sine instead of a cosine?
Remember that opposite over hypotenuse equals the sine, so
*b*/*c* = sin *B*. Multiply
through by *c* and you have
*b* = *c* × sin *B*.

Can you see how to write down two expressions for the length of side
*a*? Please work from the definitions and verify that
*a* = *c* × sin *A* =
*c* × cos *B*.

**Example:** Given, a right triangle with angle *A* =52°
and hypotenuse *c* = 150 m. What is the length of side
*b*? Hint: draw a picture, and label *A*, *c*,
and *b*.

**Solution:** Pictures are always good. You don’t
have to obsess over getting the picture exactly right, but at least
make it close. That will help you see when your answer is impossible,
so you know you’ve made a mistake. In my little sketch, I set
out to make angle *A* a bit more than 45°, but to my eyes
it looks like a bit less. That’s okay.

You may notice that I marked side *a*, even though we
don’t need it for the problem. I did that so I didn’t have
to think about which side was *b*. Always remember the rule
that the side with a given letter is opposite the angle with that
letter. (And, conventionally, we always put *C* at the right
angle, so that makes *c* the hypotenuse.)

Once you have the picture, solving the problem is pretty
straightforward. You want something involving *A*, its
adjacent side, and the hypotenuse; that has to be the cosine.

cos *A* = *b*/*c*

*b* =
*c* × cos *A* =
150 × cos 52° = about 92.35 m.

**Example:** A guy wire is anchored in the ground
and attached to the top of a 45-foot flagpole. If it meets the ground
at an angle of 63°, how long is the guy wire?

**Solution:** Presumably the flagpole is vertical, so
this is a right triangle, with
*A* = 63°, *a* = 45 ft,
and hypotenuse *c*
unknown. Which function involves the opposite side and the hypotenuse?
It must be the sine. You know that

sin *A* = *a*/*c*

Therefore,

*c* = *a*/sin *A* = 45/sin 63° =
about 50.5 ft.

You may be wondering how to find sides or angles of triangles when there is no right angle. We’ll get to that, under the topic of Solving Triangles.

One important special case comes up frequently.
Suppose the hypotenuse *c* = 1; then we call the triangle a
**unit right triangle**.
You can see from the paragraphs just above that if *c* = 1 then
*a* = sin *A* and
*b* = cos *A*.
In other words, in a
unit right triangle the opposite side will equal the sine and the
adjacent side will equal the cosine of the angle.

The triangle is often drawn
in a **unit circle**, a circle of
radius 1, as shown at right.
The angle *A* is at the center of the circle, and the adjacent side
lies along the x axis.
If you do this, the hypotenuse is the radius, which is 1.
The (*x*, *y*) coordinates of the outer end of the hypotenuse
are the *x* and *y* legs of the triangle:
(*x*, *y*) = (cos *A*, sin *A*).
**The unit circle is your friend**: it can help you visualize lots
of trig identities.

The other four functions have no real independent life of their own;
they’re **just combinations of the first two**. You could do all of
trigonometry without ever knowing more than sines and cosines. But knowing
something about the other four, especially the tangent, can often save you some
steps in a calculation—and your teacher will expect you to know
about them for exams.

I find it easiest to memorize (sorry!) the definition of the tangent in terms of the sine and cosine:

(3) **Memorize:**

tan *A* = (sin *A*) / (cos *A*)

You’ll use the tangent (tan) function a lot more than the last three functions. (I’ll get to them in a minute.)

There’s an **alternative way to remember the meaning of the tangent**.
Remember from the diagram
that sin *A* = opposite/hypotenuse and cos *A* =
adjacent/hypotenuse. Plug those into equation 3, the definition of
the tan function, and you have tan *A* =
(opposite/hypotenuse) / (adjacent/hypotenuse) or

(4) tangent = (opposite side) / (adjacent side)

Notice this is *not* marked “memorize”: you don’t have to
memorize it because it flows directly from the definition
equation 3, and in fact the two statements are
equivalent. I’ve chosen to present them in this order to minimize the
jumble of opp, adj, and hyp among sin, cos, and
tan. However, if you prefer you can memorize
equation 4 and then derive the equivalent identity
equation 3 whenever you need it.

**Example:** A guy wire is anchored in the ground and attached
to the top of a 45-foot flagpole. How far is the anchor from the base
of the flagpole, if the wire meets the ground at an
angle of 63°?

**Solution:** This is a variation on
the previous example. This
time, you want to know the side adjacent to angle *A*, not the
hypotenuse. As before, assume the flagpole is vertical, so
this is a right triangle, with
*A* = 63°, *a* = 45 ft,
and adjacent side *b*
unknown. Which function involves the adjacent side and the opposite side?
It’s the tangent. You know that

tan *A* = *a*/*b*

Therefore,

*b* = *a*/tan *A* = 45/tan 63° =
about 22.9 ft.

Now, I said you could do all of trig with just sines and cosines. How would that work for this problem? Well, sine and cosine both need the hypotenuse, so you’d have

sin *A* = *a*/*c* ⇒
*c* = *a*/sin *A* and

cos *A* = *b*/*c* ⇒
*c* = *b*/cos *A*. Therefore,

*b*/cos *A* = *a*/sin *A*

*b* = *a* × cos *A*/sin *A* =
45 × cos 63°/sin 63° = about
22.9 ft.

You got to the same place in the end, but the journey was longer. So, although it’s not strictly necessary, the tangent can make your work easier.

The other three trig
functions—**cotangent, secant, and cosecant**—are
**defined in terms of the first three**.
They’re less often used, but they do simplify some problems in
calculus. In practical problems, not involving calculus, you’ll pretty much never need them.

(5) **Memorize:**

cot *A* = 1 / (tan *A*)

sec *A* = 1 / (cos *A*)

csc *A* = 1 / (sin *A*)

Guess what! That’s the last trig identity you have to memorize.

(You’ll probably find that you end up memorizing certain other identities without even intending to, just because you use them frequently. But equation 5 makes the last ones that you’ll have to sit down and make a point of memorizing just on their own.)

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Unfortunately, the definitions in equation 5 aren’t the
easiest thing in the world to remember. Does the secant equal 1 over
the sine or 1 over the cosine? Here are **two helpful hints**:
Each of those definitions has a cofunction on one and only one side
of the equation, so you won’t be tempted to think that sec *A*
equals 1/sin *A*. And secant and cosecant go together just like
sine and cosine, so you won’t be tempted to think that cot *A*
equals 1/sin *A*.

For an alternative approach to remembering the above identities, you might like:

- This 24-second video, suggested by Gene Laratonda.
- This U-shaped approach by Bob Allison (“Robert A.”).

You can immediately notice an important relation between tangent and cotangent. Each is the cofunction of the other, just like sine and cosine:

(6) tan *A* = cot(90° − *A*) or cot(π/2 − *A*)

cot *A* = tan(90° − *A*) or tan(π/2 − *A*)

If you want to prove this, it’s easy from the definitions and equation 2:

cot *A* = 1 / tan *A*

Apply the definition of tan:

cot *A* = 1 / (sin *A* / cos *A*)

Simplify the fraction:

cot *A* = cos *A* / sin *A*

Apply equation 2:

cot *A* = sin(90° − *A*) / cos(90° − *A*)

Finally, recognize that this fraction fits the definition of the tan function, equation 3:

cot *A* = tan(90° − *A*)

Tangent and
**co**tangent are cofunctions just like sine and **co**sine.
By doing the same sort of substitution, you can show that secant and
**co**secant are also cofunctions:

(7) sec *A* = csc(90° − *A*) or csc(π/2 − *A*)

csc *A* = sec(90° − *A*) or sec(π/2 − *A*)

You saw earlier how the sine and cosine of an angle are the sides of a triangle in a unit circle. It turns out that all six functions can be shown geometrically in this way.

unit circle (radius = AB = 1)

sin θ = BC;
cos θ = AC;
tan θ = ED

csc θ = AG;
sec θ = AE;
cot θ = FG

Graphic courtesy of TheMathPage

In the illustration at right, triangle ABC has angle *θ* at the
center of a unit circle (AB = radius = 1). You already
know that BC = sin *θ* and AC = cos *θ*.

What about tan *θ*? Well, since DE is tangent to the unit
circle, you might guess that its length is tan *θ*, and you’d be
right. Triangles ABC and AED are similar, and therefore

ED / AD = BC / AC

ED / 1 = sin *θ* / cos *θ*

ED = tan *θ*

More information comes from the same pair of similar triangles:

AE / AB = AD / AC

AE / 1 = 1 / cos *θ*

AE = sec *θ*

The lengths that represent cot *θ* and csc *θ* will come
from the other triangle, GAF. That triangle is also similar to
triangle AED.
(Why? FG is perpendicular to FA, and FA is perpendicular to
AD; therefore FG and AD are parallel. In beginning geometry you
learned that when parallel lines are cut by a third line, the
corresponding angles—marked *θ* in the
diagram—are equal. Thus FG is a tangent to the unit
circle, and therefore angles *G* and *θ* are equal. )

Using similar triangles GAF and AED,

FG / FA = AD / ED

FG / 1 = 1 / tan *θ*

FG = cot *θ*

That makes sense: FG is tangent to the unit circle, and is the
tangent of the complement of angle *θ*, namely angle GAF.
Therefore, FG
is the cotangent of the original angle *θ* (or angle GAD).

Finally, using the same pair of similar triangles again, you can also say that

AG / FA = AE / ED

AG / 1 = sec *θ* / tan *θ*

AG = ( 1 / cos *θ* ) / ( sin *θ* / cos *θ* )

AG = 1 / sin *θ*

AG = csc *θ*

This one diagram beautifully depicts the geometrical meaning of all six trig functions when the angle θ is drawn at the center of a unit circle:

sin *θ* = BC;
cos *θ* = AC;
tan *θ* = ED

csc *θ* = AG;
sec *θ* = AE;
cot *θ* = FG

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.

1
Find all six functions of the angle 30°. Find sine, cosine,
and tangent of 60°.

2
Find sin *A*, sin *B*, tan *A*, and tan *B*.

3
*A* ≈ 53.13°. Find the approximate area of the
triangle. Hint: the area of a triangle is
*base* × *height*/2.

From the picture, it’s obvious why the name “tangent” makes sense: the tangent of an angle is the length of a segment tangent to the unit circle. But what about the sine function? How did it get its name?

Please look at the
picture again, and notice that
sin *θ* = BC is half a chord of the circle.
The Hindu mathematician Aryabhata the elder (about
475–550) used the word “jya” or “jiva” for this half-chord.
In Arabic translation the word was unchanged, but in the Arabic system
of writing “jiva” was written the same way as the Arabic word “jaib”,
meaning bosom, fold, or bay. The Latin word for bosom, bay, or curve
is “sinus”, or “sine” in English, and beginning with Gherardo of
Cremona (about 1114–1187) that became the standard term.

Edmund Gunter (1581–1626) seems to have been the first
to publish the abbreviations *sin* and *tan* for sin and
tangent.

My source for this history is Eli Maor’s Trigonometric Delights (1998, Princeton University Press), pages 35–36. I urge you to consult the book for a fuller account.

**27 Sept 2017**: Correct “29.2 ft” to 22.9 ft, here and here, thanks to Ryan McParlan.**29 Oct 2016**:- Added sketches for the examples of sines and cosines, and rewrote the solutions.
- Added a new example for the tangent.
- Added practice problems.
- Relabeled Why Call It Sine? as BTW.
- Updated the mathematical notation, particularly the use of italics and spaces, to conform to the standard. I used Jukka Korpela’s comprehensive Writing Mathematical Expressions (2014, Suomen E-painos Oy), ISBN 978-952-6613-25-3.

- (intervening changes suppressed)
**19 Feb 1997**: New document.

next: 3/Special Angles

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