64 2. THE LIBRARY OF FUNCTIONS
1
-1
0
2
3
2
2
2
3
2
2
Figure 2.6: e sine (light) and cosine (dark) functions.
Knowledge Box 2.20
Periodicity identities
• sin.x C 2/ D sin.x/
• cos.x C 2/ D cos.x/
• sin.x/ D cos
x
2
• tan.x/ D cot
x
2
• sec.x/ D csc
x C
2
• cos.x/ D cos.x/
• sin.x/ D sin.x/
• tan.x/ D tan.x/
• sin.x C / D sin.x/
• cos.x C / D cos.x/
• tan.x C / D tan.x/
2.3.2 THEOREMS ABOUT TRIANGLES
e most basic fact about triangles is that the sum of the angles of a triangle in the plane is
radians. is means that if we know two of the angles, we can recover the third by taking
minus their sum.
If we have a right triangle with a hypotenuse of length c and legs of length a and b, then the
Pythagorean theorem tells us that
a
2
C b
2
D c
2
:
Homework problem 2.69 asks you to show that the fact
sin
2
./ C cos
2
D 1
is an instance of the Pythagorean theorem. In fact, there are several useful relations between the
trigonometric functions that arise from this fact.
2.3. TRIGONOMETRIC FUNCTIONS 65
1
-1
0
2
3
2
2
3
2
2
Figure 2.7: e tangent (light) and cotangent (dark) functions.
1
-1
0
2
3
2
2
3
2
2
Figure 2.8: e secant (light) and cosecant (dark) functions.
66 2. THE LIBRARY OF FUNCTIONS
Knowledge Box 2.21
e Pythagorean identities
For any angle , we have:
• sin
2
./ C cos
2
./ D 1
• tan
2
./ C 1 D sec
2
./
• 1 C cot
2
./ D csc
2
./
ere are some handy relationships that apply to the sides and angles of all triangles (not just
right triangles). ese are phrased in terms of the general triangle shown in Figure 2.9. Both of
these are used to solve problems involving arbitrary triangles. ey are called the law of sines
and the law of cosines.
˛
ˇ
A
B
C
Figure 2.9: A general triangle with labeled side lengths and angles.
Knowledge Box 2.22
e law of sines
A
sin.˛/
D
B
sin.ˇ/
D
C
sin./
2.3. TRIGONOMETRIC FUNCTIONS 67
Knowledge Box 2.23
e law of cosines
C
2
D A
2
C B
2
C 2AB cos./
Example 2.51 Suppose for the triangle shown in Figure 2.9 we have A D 3, B D 5, and
ˇ D
7
. What are ˛, , and C ?
Solution:
Using the law of sines:
A
sin.˛/
D
B
sin.ˇ/
3
sin.˛/
D
5
sin.=7/
sin.˛/ D
3 sin.=7/
5
˛ D sin
1
3 sin.=7/
5
˛ Š 0:273 rad
For the moment, treat sin
1
as a button on your calculator. We will get to inverse trig functions
in Section 3.3.3.
We know that ˛ C ˇ C D , so
D
7
0:273 Š 2:42 rad:
Now that we know
, we can apply the law of sines again to get
C
:
C D
sin./ B
sin
.ˇ/
Š 7:61
˙
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