Download Trigonometric Functions and Triangles

Trigonometric Functions and Triangles
Dr. Philippe B. Laval
Kennesaw STate University
September 2, 2012
Abstract
These notes describe techniques which allow us to …nd all the trigonometric functions knowing one of them, by using a triangle. We apply
these techniques to …nd the value of expressions of the form tan sin 1 a ,
sin cos 1 a , cos tan 1 a , ...for a given value of a.
1
Trigonometric Functions and Triangles
In order to understand the material in these notes, it is important to remember
how trigonometric functions can be expressed in terms of the sides of a right
triangle. We reviews these formulas. Consider the triangle in …gure 1. Then,
1.1
sin t
=
cos t
=
tan t
=
cot t
=
sec t
=
csc t
=
a
opposite side
=
hypotenuse
c
b
adjacent side
=
hypotenuse
c
sin t
opposite side
a
=
=
cos t
adjacent side
b
1
cos t
adjacent side
b
=
=
=
tan t
sin t
opposite side
a
1
hypotenuse
c
=
=
cos t
adjacent side
b
1
hypotenuse
c
=
=
sin t
opposite side
a
Finding Trigonometric Functions Knowing One of Them
The goal here is knowing one of the trigonometric functions, how do we …nd
the remaining …ve trigonometric functions. There are di¤erent ways for doing
this, including using trigonometric identities or a calculator. The technique
we present here involves using right triangles. It is very simple and has the
1
Figure 1: Right Triangle
advantage of giving an exact answer. This method also has a nice application
as we see in the next section. We explain the technique by examples.
2
and sin t = , …nd cos t and tan t.
2
3
2
We draw a right triangle in which sin t = . This is easy, From above, we
3
2
opposite side
have sin t =
= . This, the triangle that will work is a right
hypotenuse
3
triangle where the side opposite the angle t is 2 units long and the hypotenuse is
3 units long (see …gure 2). Let x be the length of the remaining side (adjacent
side). We can …nd it usingpthe Pythagorean theorem. We have x2 + 22 = 32 or
x2 = 9 4 = 5. Thus x = 5. Hence
p
adjacent side
5
=
cos t =
hypotenuse
3
Example 1 Suppose that 0 < t <
and
tan t =
1.2
opposite side
2
=p
adjacent side
5
Composing Trigonometric Functions and their Inverses
It is often important to be able to evaluate expressions of the form tan sin 1 a ,
sin cos 1 a , cos tan 1 a , ...for a given value of a. This is the case in particular
in calculus. This can be done with a calculator if we know the value of a.
There is also an elegant way to do it using triangles. This way gives a very
accurate answer (exact most of the time), while the calculator will only give an
approximation. We illustrate it with an example.
2
Figure 2: Right Triangle
3
.
5
We begin by looking at the inverse trigonometric function. We say, let t =
3
3
cos 1 . Then, sin cos 1
= sin t. So, we must …nd sin t. By de…nition,
5
5
3
3
if t = cos 1 , then cos t = . Therefore the problem becomes to …nd sin t,
5
5
knowing cos t. We know how to do this, we did it above. This is the problem of
…nding all the trigonometric ratios, knowing one of them. We begin by drawing
3
a triangle in which cos t = . Such a triangle is shown in …gure 3. Then, in
5
x
this triangle sin t = , so we need to …nd x. Using the Pythagorean theorem,
5
we have
Example 2 Find sin cos
Therefore, sin t =
1
x2 + 9
=
25
x
=
4
3
5
=
4
. Hence,
5
sin cos
1.3
1
4
5
Problems
All problems below must be dine using the techniques described in this document.
1. Given that cos t =
1
, …nd the …ve remaining trigonometric functions.
5
3
Figure 3: Right Triangle
2. Given that tan t =
p
3
, …nd the …ve remaining trigonometric functions.
3
3. Given that sec t = 3, …nd the …ve remaining trigonometric functions.
4. Find tan sin
1
5. Find cos tan
1
1.4
1
2
1
3
Answers
1. Given that cos t =
1
, …nd the …ve remaining trigonometric functions.
5
p
24
sin t =
5
1
cos t =
5
p
24
tan t =
1
cot t = p
24
sec t = 5
5
csc t = p
24
4
2. Given that tan t =
p
3
, …nd the …ve remaining trigonometric functions.
3
sin t =
cos t =
tan t =
cot t =
sec t =
csc t =
1
2p
3
p2
3
3
p
3
p
2 3
3
2
3. Given that sec t = 3, …nd the …ve remaining trigonometric functions.
p
2 2
sin t =
3
1
cos t =
3p
tan t = 2 2
p
2
cot t =
4
sec t = 3 p
3 2
csc t =
4
4. Find tan sin
1
1
2
1
tan sin
5. Find cos tan
1
1
2
=
1
3
cos tan
5
1
1
3
=
p
3
3
p
3 10
10