B
B
5
3
B′
B′
5
3
A
A
4
C
A′
4
C′
A′
C′
C
(a) Inches
(b) Centimeters
(c) Similar triangles
Figure 1.2.1
△ ABC ∼ △ A′ B′ C′
If the American triangle is △ ABC and the German triangle is △ A′ B′ C′, then we see
from Figure 1.2.1 that △ ABC is similar to △ A′ B′ C′, and hence the corresponding angles Trigonometric Functions of an Acute Angle • Section 1.2
9
are equal and the ratios of the corresponding sides are equal. In fact, we know that com-
mon ratio: the sides of △ ABC are approximately 2.54 times longer than the corresponding
sides of △ A′ B′ C′. So when the American student calculates sin A and the German student
calculates sin A′, they get the same answer:3
BC
AB
BC
B′ C′
△ ABC ∼ △ A′ B′ C′
⇒
=
⇒
=
⇒
sin A = sin A′
B′ C′
A′ B′
AB
A′ B′
Likewise, the other values of the trigonometric functions of A and A′ are the same. In fact,
our argument was general enough to work with any similar right triangles. This leads us to
the following conclusion:
When calculating the trigonometric functions of an acute angle A, you may
use any right triangle which has A as one of the angles.
Since we defined the trigonometric functions in terms of ratios of sides, you can think
of the units of measurement for those sides as canceling out in those ratios. This means
that the values of the trigonometric functions are unitless numbers. So when the American
student calculated 3/5 as the value of sin A in Example 1.5, that is the same as the 3/5 that
the German student calculated, despite the different units for the lengths of the sides.
Example 1.6
Find the values of all six trigonometric functions of 45◦.
1
B
Solution: Since we may use any right triangle which has 45◦ as one of the
angles, use the simplest one: take a square whose sides are all 1 unit long and
2
1
1
divide it in half diagonally, as in the figure on the right. Since the two legs
of the triangle △ ABC have the same length, △ ABC is an isosceles triangle,
45◦
which means that the angles A and B are equal. So since A + B = 90◦, this
A
1
C
means that we must have A = B = 45◦. By the Pythagorean Theorem, the
length c of the hypotenuse is given by
c 2 = 12 + 12 = 2
⇒
c =
2 .
Thus, using the angle A we get:
opposite
1
adjacent
1
opposite
1
sin 45◦ =
=
cos 45◦ =
=
tan 45◦ =
=
= 1
hypotenuse
2
hypotenuse
2
adjacent
1
hypotenuse
hypotenuse
adjacent
1
csc 45◦ =
=
2
sec 45◦ =
=
2
cot 45◦ =
=
= 1
opposite
adjacent
opposite
1
Note that we would have obtained the same answers if we had used any right triangle similar to
△ ABC. For example, if we multiply each side of △ ABC by
2, then we would have a similar
triangle with legs of length
2 and hypotenuse of length 2. This would give us sin 45◦ = 2 , which
2
equals
2
= 1 as before. The same goes for the other functions.
2· 2
2
3We will use the notation AB to denote the length of a line segment AB.
10
Chapter 1 • Right Triangle Trigonometry
§1.2
Example 1.7
Find the values of all six trigonometric functions of 60◦.
B
Solution: Since we may use any right triangle which has 60◦ as one of
the angles, we will use a simple one: take a triangle whose sides are all 2
units long and divide it in half by drawing the bisector from one vertex to
2
30◦
2
the opposite side, as in the figure on the right. Since the original triangle
3
was an equilateral triangle (i.e. all three sides had the same length), its
three angles were all the same, namely 60◦. Recall from elementary ge-
60◦
60◦
ometry that the bisector from the vertex angle of an equilateral triangle
A
1
C
1
to its opposite side bisects both the vertex angle and the opposite side. So
as in the figure on the right, the triangle
2
△ ABC has angle A = 60◦ and
angle B = 30◦, which forces the angle C to be 90◦. Thus, △ ABC is a right
triangle. We see that the hypotenuse has length c = AB = 2 and the leg AC has length b = AC = 1.
By the Pythagorean Theorem, the length a of the leg BC is given by
a 2 + b 2 = c 2
⇒
a 2 = 22 − 12 = 3
⇒
a =
3 .
Thus, using the angle A we get:
opposite
3
adjacent
1
opposite
3
sin 60◦ =
=
cos 60◦ =
=
tan 60◦ =
=
=
3
hypotenuse
2
hypotenuse
2
adjacent
1
hypotenuse
2
hypotenuse
adjacent
1
csc 60◦ =
=
sec 60◦ =
= 2
cot 60◦ =
=
opposite
3
adjacent
opposite
3
Notice that, as a bonus, we get the values of all six trigonometric functions of 30◦, by using angle
B = 30◦ in the same triangle △ ABC above:
opposite
1
adjacent
3
opposite
1
sin 30◦ =
=
cos 30◦ =
=
tan 30◦ =
=
hypotenuse
2
hypotenuse
2
adjacent
3
hypotenuse
hypotenuse
2
adjacent
3
csc 30◦ =
= 2
sec 30◦ =
=
cot 30◦ =
=
=
3
opposite
adjacent
3
opposite
1
Example 1.8
A is an acute angle such that sin A = 2 . Find the values of the other trigonometric
B
3
functions of A.
3
2
Solution: In general it helps to draw a right triangle to solve problems of this
type. The reason is that the trigonometric functions were defined in terms of
A
b
C
ratios of sides of a right triangle, and you are given one such function (the sine,
in this case) already in terms of a ratio: sin A = 2 . Since sin A is defined as
3
opposite , use 2 as the length of the side opposite A and use 3 as the length of the hypotenuse in a
hypotenuse
right triangle △ ABC (see the figure above), so that sin A = 2 . The adjacent side to A has unknown
3
length b, but we can use the Pythagorean Theorem to find it:
22 + b 2 = 32
⇒
b 2 = 9 − 4 = 5
⇒
b =
5
Trigonometric Functions of an Acute Angle • Section 1.2
11
We now know the lengths of all sides of the triangle △ ABC, so we have:
adjacent
5
opposite
2
cos A =
=
tan A =
=
hypotenuse
3
adjacent
5
hypotenuse
3
hypotenuse
3
adjacent
5
csc A =
=
sec A =
=
cot A =
=
opposite
2
adjacent
5
opposite
2
You may have noticed the connections between the sine and cosine, secant and cosecant,
and tangent and cotangent of the complementary angles in Examples 1.5 and 1.7. Generalizing those examples gives us the following theorem:
Theorem 1.2. Cofunction Theorem: If A and B are the complementary acute angles in a
right triangle △ ABC, then the following relations hold:
sin A = cos B
sec A = csc B
tan A = cot B
sin B = cos A
sec B = csc A
tan B = cot A
We say that the pairs of functions { sin, cos }, { sec, csc }, and { tan, cot } are cofunctions.
So sine and cosine are cofunctions, secant and cosecant are cofunctions, and tangent and
cotangent are cofunctions. That is how the functions cosine, cosecant, and cotangent got the
“co” in their names. The Cofunction Theorem says that any trigonometric function of an
acute angle is equal to its cofunction of the complementary angle.
Example 1.9
Write each of the following numbers as trigonometric functions of an angle less than 45◦: (a) sin 65◦;
(b) cos 78◦; (c) tan 59◦.
Solution: (a) The complement of 65◦ is 90◦ − 65◦ = 25◦ and the cofunction of sin is cos, so by the
Cofunction Theorem we know that sin 65◦ = cos 25◦.
(b) The complement of 78◦ is 90◦ − 78◦ = 12◦ and the cofunction of cos is sin, so cos 78◦ = sin 12◦.
(c) The complement of 59◦ is 90◦ − 59◦ = 31◦ and the cofunction of tan is cot, so tan 59◦ = cot 31◦.
2 a
a 2
60◦
45◦
a
a
30◦
45◦
a
a 3
(a) 45−45−90
(b) 30 − 60 − 90
Figure 1.2.2
Two general right triangles (any a > 0)
The angles 30◦, 45◦, and 60◦ arise often in applications. We can use the Pythagorean
Theorem to generalize the right triangles in Examples 1.6 and 1.7 and see what any 45 −
45 − 90 and 30 − 60 − 90 right triangles look like, as in Figure 1.2.2 above.
12
Chapter 1 • Right Triangle Trigonometry
§1.2
Example 1.10
Find the sine, cosine, and tangent of 75◦.
D
Solution: Since 75◦ = 45◦+30◦, place a 30−60−90 right triangle
△ ADB with legs of length
3 and 1 on top of the hypotenuse
of a 45 − 45 − 90 right triangle △ ABC whose hypotenuse has
1
length
3, as in the figure on the right. From Figure 1.2.2(a) we
know that the length of each leg of △ ABC is the length of the
B
hypotenuse divided by
2. So AC = BC = 3 =
3 . Draw DE
F
2
2
perpendicular to AC, so that △ ADE is a right triangle. Since
2
∠ BAC = 45◦ and ∠ D AB = 30◦, we see that ∠ D AE = 75◦ since
3
it is the sum of those two angles. Thus, we need to find the sine,
3
cosine, and tangent of ∠ D AE.
2
Notice that ∠ ADE = 15◦, since it is the complement of ∠ D AE.
And ∠ ADB = 60◦, since it is the complement of ∠ D AB. Draw
◦
30
BF perpendicular to DE, so that △ DFB is a right triangle.
Then ∠ BDF = 45◦, since it is the difference of ∠ ADB = 60◦ and
45◦
E
A
C
∠ ADE = 15◦. Also, ∠ DBF = 45◦ since it is the complement of
3
2
∠ BDF. The hypotenuse BD of △ DFB has length 1 and △ DFB
is a 45 − 45 − 90 right triangle, so we know that DF = FB = 1 .
2
Now, we know that DE ⊥ AC and BC ⊥ AC, so FE and BC are parallel. Likewise, FB and EC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since ∠ BCE
is a right angle. So EC = FB = 1 and FE = BC =
3 . Hence,
2
2
DE = DF + FE = 1 +
3
and
AE = AC − EC =
3
=
3 − 1 . Thus,
2
2 =
3 + 1
2
2 −
1
2
2
3+1
3−1
3+1
sin 75◦ = DE
2
, cos 75◦
2
, and tan 75◦
2
.
AD =
2
= 6+ 2
4
= AE
AD =
2
= 6− 2
4
= DE
AE =
= 6+ 2
3−1
6− 2
2
Note: Taking reciprocals, we get csc 75◦ =
4
, sec 75◦ =
4
, and cot 75◦ = 6− 2 .
6+ 2
6− 2
6+ 2
Exercises
For Exercises 1-10, find the values of all six trigonometric functions of
B
angles A and B in the right triangle △ ABC in Figure 1.2.3.
c
a
1. a = 5, b = 12, c = 13
2. a = 8, b = 15, c = 17
A
b
C
3. a = 7, b = 24, c = 25
4. a = 20, b = 21, c = 29
Figure 1.2.3
5. a = 9, b = 40, c = 41
6. a = 1, b = 2, c = 5
7. a = 1, b = 3
8. a = 2, b = 5
9. a = 5, c = 6
10. b = 7, c = 8
For Exercises 11-18, find the values of the other five trigonometric functions of the acute angle A
given the indicated value of one of the functions.
Trigonometric Functions of an Acute Angle • Section 1.2
13
11. sin A = 3
12. cos A
13. cos A
14. sin A
4
= 23
= 2
= 2
10
4
15. tan A = 5
16. tan A
18. csc A
9
= 3
17. sec A = 73
= 3
For Exercises 19-23, write the given number as a trigonometric function of an acute angle less than
45◦.
19. sin 87◦
20. sin 53◦
21. cos 46◦
22. tan 66◦
23. sec 77◦
For Exercises 24-28, write the given number as a trigonometric function of an acute angle greater
than 45◦.
24. sin 1◦
25. cos 13◦
26. tan 26◦
27. cot 10◦
28. csc 43◦
29. In Example 1.7 we found the values of all six trigonometric functions of 60◦ and 30◦.
(a) Does sin 30◦ + sin 30◦ = sin 60◦?
(b) Does cos 30◦ + cos 30◦ = cos 60◦?
(c) Does tan 30◦ + tan 30◦ = tan 60◦?
(d) Does 2 sin 30◦ cos 30◦ = sin 60◦?
30. For an acute angle A, can sin A be larger than 1? Explain your answer.
31. For an acute angle A, can cos A be larger than 1? Explain your answer.
32. For an acute angle A, can sin A be larger than tan A? Explain your answer.
33. If A and B are acute angles and A < B, explain why sin A < sin B.
34. If A and B are acute angles and A < B, explain why cos A > cos B.
35. Prove the Cofunction Theorem (Theorem 1.2). ( Hint: Draw a right triangle and label the angles
and sides. )
36. Use Example 1.10 to find all six tr