Transcript trigonometric functions

6. The Trigonometric Functions
6.2 Trigonometric functions of angles
Copyright © Cengage Learning. All rights reserved.
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Trigonometric Functions of Angles
A triangle is a right triangle if one of its angles is a right
angle. If  is any acute angle, we may consider a right
triangle having  as one of its angles, as in Figure 1, where
the symbol
specifies the 90° angle.
Figure 1
Six ratios can be obtained using the lengths a, b, and c of
the sides of the triangle:
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Trigonometric Functions of Angles
We can show that these ratios depend only on , and not
on the size of the triangle, as indicated in Figure 2.
Figure 2
Since the two triangles have equal angles, they are similar,
and therefore ratios of corresponding sides are
proportional.
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Trigonometric Functions of Angles
For example,
Thus, for each , the six ratios are uniquely determined and
hence are functions of .
They are called the trigonometric functions and are
designated as the sine, cosine, tangent, cotangent,
secant, and cosecant functions, abbreviated sin, cos,
tan, cot, sec, and csc, respectively.
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Trigonometric Functions of Angles
The symbol sin( ), or sin , is used for the ratio b/c, which
the sine function associates with . Values of the other five
functions are denoted in similar fashion.
To summarize, if  is the acute angle of the right triangle in
Figure 1, then, by definition,
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Trigonometric Functions of Angles
The domain of each of the six trigonometric functions is the
set of all acute angles. Later in this section we will extend
the domains to larger sets of angles, and in the next
section, to real numbers.
If  is the angle in Figure 1, we refer to the sides of the
triangle of lengths a, b, and c as the adjacent side,
opposite side, and hypotenuse, respectively.
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Trigonometric Functions of Angles
We shall use adj, opp, and hyp to denote the lengths of
the sides. We may then represent the triangle as in
Figure 3.
Figure 3
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Trigonometric Functions of Angles
With this notation, the trigonometric functions may be
expressed as follows.
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Trigonometric Functions of Angles
Note that since
sin  and csc  are reciprocals of each other, giving us the
two identities in the left-hand column of the next box.
Similarly, cos  and sec  are reciprocals of each other, as
are tan  and cot .
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Example 1 – Finding trigonometric function values
If  is an acute angle and cos  =
trigonometric functions of .
, find the values of the
Solution:
We begin by sketching a right triangle having an acute
angle  with adj = 3 and hyp = 4, as shown in Figure 4, and
proceed as follows:
Figure 4
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Example 1 – Solution
32 + (opp)2 = 42
(opp)2 = 16 – 9
cont’d
Pythagorean theorem
isolate (opp)2
=7
opp =
take the square root
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Example 1 – Solution
cont’d
Applying the definition of the trigonometric functions of an
acute angle of a right triangle, we obtain the following:
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Example 2 – Finding trigonometric function values of 60°, 30°, and 45°
Find the values of the trigonometric functions that
correspond to  :
(a)  = 60°
(b)  = 30°
(c)  = 45°
Solution:
Consider an equilateral triangle with
sides of length 2. The median from
one vertex to the opposite side
bisects the angle at that vertex,
as illustrated by the dashes in
Figure 5.
Figure 5
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Example 2 – Solution
cont’d
By the Pythagorean theorem, the side opposite 60° in the
shaded right triangle has length
.
Using the formulas for the trigonometric functions of an
acute angle of a right triangle, we obtain the values
corresponding to 60° and 30° as follows:
(a)
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Example 2 – Solution
cont’d
(b)
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Example 2 – Solution
cont’d
(c) To find the values for  = 45°, we may consider an
isosceles right triangle whose two equal sides have
length 1, as illustrated in Figure 6.
Figure 6
By the Pythagorean theorem, the length of the
hypotenuse is
.
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Example 2 – Solution
cont’d
Hence, the values corresponding to 45° are as follows:
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Trigonometric Functions of Angles
Because of the importance of these special values, it is a
good idea either to memorize the table or to learn to find
the values quickly by using triangles.
Special Values of the Trigonometric Functions
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Trigonometric Functions of Angles
The next example illustrates a practical use for
trigonometric functions of acute angles.
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Example 3 – Finding the height of a flagpole
A surveyor observes that at a point A, located on level
ground a distance 25.0 feet from the base B of a flagpole,
the angle between the ground and the top of the pole is
30°. Approximate the height h of the pole to the nearest
tenth of a foot.
Solution:
Referring to Figure 7, we
see that we want to relate
the opposite side and the
adjacent side, h and 25,
respectively, to the 30° angle.
Figure 7
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Example 3 – Solution
cont’d
This suggests that we use a trigonometric function
involving those two sides—namely, tan or cot. It is usually
easier to solve the problem if we select the function for
which the variable is in the numerator.
Hence, we have
or, equivalently,
h = 25 tan 30.
We use the value of tan 30° from Example 2 to find h:
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Trigonometric Functions of Angles
Let us next list all the fundamental identities. These
identities are true for every acute angle , and  may take
on various forms.
We shall see later that these identities are also true for
other angles and for real numbers.
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Trigonometric Functions of Angles
We can use the fundamental identities to express each
trigonometric function in terms of any other trigonometric
function.
Two illustrations are given in the next example.
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Example 4 – Using fundamental identities
Let  be an acute angle.
(a) Express sin  in terms of cos .
(b) Express tan  in terms of sin .
Solution:
(a) We may proceed as follows:
sin2  + cos2  = 1
sin2  = 1 – cos2 
Pythagorean identity
isolate sin2 
sin  =
take the square root
sin  =
sin  > 0 for acute angles
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Example 4 – Solution
cont’d
(b) If we begin with the fundamental identity
then all that remains is to express cos  in terms of sin .
We can do this by solving sin2  + cos2  = 1 for cos ,
obtaining
Hence,
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Trigonometric Functions of Angles
Fundamental identities are often used to simplify
expressions involving trigonometric functions, as illustrated
in the next example.
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Example 5 – Showing that an equation is an identity
Show that the following equation is an identity by
transforming the left-hand side into the right-hand side:
(sec  + tan  ) (1 – sin  ) = cos 
Solution:
We begin with the left-hand side and proceed as follows:
reciprocal and
tangent identities
add fractions
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Example 5 – Solution
cont’d
multiply
sin2  + cos2  = 1
cancel cos 
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Trigonometric Functions of Angles
Since many applied problems involve angles that are not
acute, it is necessary to extend the definition of the
trigonometric functions.
We make this extension by using the standard position of
an angle  on a rectangular coordinate system.
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Trigonometric Functions of Angles
If  is acute, we have the situation illustrated in Figure 11,
Figure 11
where we have chosen a point P(x, y) on the terminal side
of  and where d(O, P) = r =
.
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Trigonometric Functions of Angles
Referring to triangle OQP, we have
We shall define the six trigonometric functions so that their
values agree with those given previously whenever the
angle is acute.
It is understood that if a zero denominator occurs, then the
corresponding function value is undefined.
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Trigonometric Functions of Angles
Our discussion of domains is summarized in the following
table, where n denotes any integer.
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Trigonometric Functions of Angles
For any point P(x, y) in the preceding definition, |x|  r and
|y|  r or, equivalently |x/r|  1 , and |y/r|  1.
Thus,
|sin  |  1, |cos  |  1, |csc  |  1, and |sec  |  1,
for every  in the domains of these functions.
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Example 8 – Finding trigonometric function values of an angle in standard position
An angle  is in standard position, and its terminal side lies
in quadrant III on the line y = 3x. Find the values of the
trigonometric functions of .
Solution:
The graph of y = 3x is
sketched in Figure 14,
together with the initial
and terminal sides of .
Figure 14
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Example 8 – Solution
cont’d
Since the terminal side of  is in quadrant III, we begin by
choosing a convenient negative value of x, say x = –1.
Substituting in y = 3x gives us y = 3(–1) = –3, and hence
P(–1, –3) is on the terminal side. Applying the definition of
the trigonometric functions of any angle with x = –1, y = –3,
and
gives us
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Trigonometric Functions of Angles
The definition of the trigonometric functions of any angle
may be applied if  is a quadrantal angle. The procedure is
illustrated by the next example.
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Example 9 – Finding trigonometric function values of a quadrantal angle
If  = 3 /2, find the values of the trigonometric functions
of .
Solution:
Note that 3 /2 = 270. If  is
placed in standard position,
the terminal side of  coincides
with the negative y-axis, as
shown in Figure 15.
Figure 15
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Example 9 – Solution
cont’d
To apply the definition of the trigonometric functions of any
angle, we may choose any point P on the terminal side
of . For simplicity, we use P(0, –1).
In this case, x = 0, y = –1, r = 1, and hence
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Example 9 – Solution
cont’d
The tangent and secant functions are undefined, since the
meaningless expressions tan  = (–1)/0 and sec  = 1/0
occur when we substitute in the appropriate formulas.
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Trigonometric Functions of Angles
Let us determine the signs associated with values of the
trigonometric functions.
If  is in quadrant II and P(x, y) is a point on the terminal
side, then x is negative and y is positive.
Hence, sin  = y/r and csc  = r/y are positive, and the
other four trigonometric functions, which all involve x, are
negative.
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Trigonometric Functions of Angles
Checking the remaining quadrants in a similar fashion, we
obtain the following table.
Signs of the Trigonometric Functions
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Trigonometric Functions of Angles
The diagram in Figure 16 may be useful for remembering
quadrants in which trigonometric functions are positive. If a
function is not listed (such as cos in quadrant II), then that
function is negative.
Positive trigonometric functions
Figure 16
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Example 10 – Finding the quadrant containing an angle
Find the quadrant  containing if both cos  > 0 and
sin  < 0.
Solution:
Referring to the table of signs or Figure 16, we see that
cos  > 0 (cosine is positive) if  is in quadrant I or IV and t
that sin  < 0 (sine is negative) if  is in quadrant III or IV.
Hence, for both conditions to be satisfied , must be in
quadrant IV.
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Example 11 – Finding values of trigonometric functions from prescribed conditions
If sin  = and tan  < 0, use fundamental identities to find
the values of the other five trigonometric functions.
Solution:
Since sin  = > 0 (positive) and tan  < 0 (negative),  is
in quadrant II.
Using the relationship sin2  + cos2  = 1 and the fact that
cos  is negative in quadrant II, we have
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Example 11 – Solution
cont’d
Next we use the tangent identity to obtain
Finally, using the reciprocal identities gives us
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Example 12 – Using fundamental identities
Rewrite
in nonradical form without
using absolute values for  <  < 2.
Solution:
cos2  + sin2  = 1
1 + cot2  = csc2 
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Example 12 – Solution
cont’d
Since  <  < 2, we know that  is in quadrant III or IV.
Thus, csc  is negative, and by the definition of absolute
value, we have
|csc  | = –csc .
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