There are a few (a very few) angles that have relatively "neat" trigonometric values,
involving, at worst, one square root. Because of their relatively simple
values, these are the angles which will typically be used in math problems
(in calculus, especially), and you will be expected to have these angles' values memorized.

Usually, textbooks present
these values in a table that you are expected memorize. But pictures are
generally easier to recall (on tests, etc); this lesson will show the
way in which many people really keep track of these values.

45°-Angle Values (from
a 45-45-90 triangle)

If you need to work
with a 45-degree
angle, then draw the 45-45-90-degree
triangle at the right, using the given values for the lengths of
the sides:

We choose a length of one
unit for the matching sides because this is simplest, and then we get
the sqrt(2) value by using the Pythagorean Theorem. (Note that it is irrelevant what
are the lengths of the actual triangle you are dealing with; this reference
triangle gives you the ratios, and thus the trigonometric values.) The
"theta" (the squashed circle with the line through it) in the
lower left corner is the angle we'll use. (The orientation of the particular
angle they want you to work with is irrelevant, because you can always
rotate the above triangle to put it into whatever orientation you need.)

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If, for a given exercise,
they only want the trigonometric value of the angle, then just read it
off the triangle: the sine of theta is (opposite) over (hypoteneuse),
or 1/sqrt(2);
the cosine of theta is (adjacent) over (hypoteneuse), or 1/sqrt(2);
and the tangent of theta is (opposite) over (adjacent), or 1/1
= 1. (Your text or
teacher might want you to "rationalize" these ratios, in which
case you get sin(theta)
= sqrt(2)/2 and cos(theta)
= sqrt(2)/2. This
rationalization issue becomes much less important once you reach calculus.)

On the other hand, suppose
they've given you a 45-degree
triangle where the two matching sides have length, say, 14 units, and they want you to find the length of the third side. Okay, you
multiply 1 by 14 to get 14 for the matching sides. Then, by similar triangles, you multiply sqrt(2) by 14 to get 14sqrt(2) for the length of the hypoteneuse.

You can find anything you
need from this reference triangle. Instead of trying to memorize an entire
table (if that's not working for you), simply memorize this triangle.

30°- and 60°-Angle
Values (from a 30-60-90 triangle)

If you need to work with
a 30-
or a 60-degree
angle, the process is similar to the above, but the set-up is a bit longer.

For either of the
angles, this is the triangle that you start with:

This is a 60-60-60 triangle (that is, an equilateral triangle), with sides having
a length of two units.

Drop
the vertical bisector from the top angle down to the bottom side:

Note that this
bisector is also the altitude (height) of the triangle.

By using the Pythagorean
Theorem, we get that the length of the bisector is sqrt(3),
and the bisector has formed 30-60-90 triangles.

If you are doing
a 60-degree
triangle, use the angle labelled above as "alpha" (the
funny-looking "a"
in the lower corner); if you are doing a 30-degree
triangle, use the angle labelled with "beta" (the funny-looking
"b"
in the upper corner).

For working with 60-degree
angles, your picture is this half of the triangle:

...and for 30-degree
angles, your picture is the same half, but rotated:

You find trigonometric
values and ratios with the 30-degree
and 60-degree
triangles in the exact same manner as with the 45-degree
triangle.

You may get one of those
teachers who doesn't want you to draw these pictures (because you're supposed
to have everything memorized by now). Well, this is why your pencil has
an eraser. My Calculus II instructor said that if we drew the pictures
on our tests, the entire problem would be counted wrong. I drew the pictures
anyway, but very lightly, and erased them all before I handed the tests
in. He never knew, and I passed the course. You do what you gotta do.

On the other hand, some
people prefer tables and charts. If tables work better for you, then this
table comes highly recommended, having been "field-tested" by
a working instructor:

30°

45°

60°

sin

1

2

3

cos

3

2

1

divided
by 2

To find, say, the sine
of forty-five degrees, you would trace across in the "sin"
row and down the "45°"
column, taking the square-root symbol with you as you go and remembering
to include the "divided by 2"
from the bottom, to get sin(45°)
= sqrt(2)/2.
The neat pattern of "1,
2, 3" across the
top row and "3,
2, 1" across the
middle row are meant to help you memorize the table values. Keep in mind
that the square root of 1 is just 1,
so, for instance, cos(60°)
= sqrt(1)/2
= 1/2.