Transcript Do 5.2

Chapter 5
Section 5.2
Trigonometric Functions
Angles in Standard Position
y-axis
Recall an angle in standard position is an angle that has its
initial side on the positive x-axis. We can use any point on the
angles terminal side to find the values of the trigonometric
ratios. If the coordinates of the point P are (x,y) and the
distance the point P is from the origin is r we get the
following values for the trigonometric ratios.
y
sin   
r
cos  
y
x2  y2
x

r
x
x2  y 2
y
tan  
x
x
cot  
y
In the example to the right with the coordinates of P at the
point (1,3)
3
sin  
10
1
cos  
10
3
tan  
1
1
cot  
3
10
sec  
1
10
csc  
3
P:(x,y)
r  x2  y2
y

x-axis
x
r
sec   
x
x2  y2
x
r
csc   
y
x2  y 2
y
P:(1,3)
3
2 r
1
r  12  32
r  10

1
2
It is important to realize that it does not matter what point you
select on the terminal side of the angle the trigonometric ratios
will be the same because the triangles are similar. The triangle
with its vertex at P1 is similar to the triangle with its vertex at P2
and the length of the sides are proportional (equal ratios).
P2
P1

Signs of Trigonometric Functions
The trigonometric ratios now are defined no matter where the terminal
side of the angle is. It can be in any if the four quadrants. Since the
values for the xy-coordinates are different signs (±) depending on the
quadrant the trigonometric ratios will be also. The value for r is always
positive. The chart below shows the signs of the trigonometric ratios.
x neg (-)
y pos (+)
x pos (+)
y pos (+)
x neg (-)
y neg (-)
x pos (+)
y neg (-)
Quadrant
sin
cos
tan
cot
sec
csc
I
+
+
+
+
+
+
II
+
-
-
-
-
+
III
-
-
+
+
-
-
IV
-
+
-
-
+
-
Find the values of the six trigonometric functions for the left side of
the line 2𝑥 + 3𝑦 = 0. The point (-3,2) is on the terminal side of the
angle. We find the value for r (distance from the origin) first.
2
r
r  (3) 2  2 2  9  4  13

-3
2
3
2
3
13
13
sin  
cos  
cot  
tan  
csc  
sec  
13
13
2
3
2
3
Any point on one side of a line can be used to determine the values of the trigonometric functions. To find the
values at the 4 angles 0° , 90° , 180° , 270° , and 360° use the 4 points pictured and keep in mind 𝑟 = 1.
Angle
x
y
sin
cos
tan
cot
sec
csc
90
0°
1
0
0
1
0
undefined
1
undefined
1
90°
180°
0.5
0
-1
1
0
1
0
0
-1
undefined
0
0
undefined
undefined
-1
1
undefined
180
1
0.5
0.5
0.5
270°
0
-1
-1
0
undefined
0
undefined
-1
1
360°
1
0
0
1
0
undefined
1
undefined
270
1
0 ,360
Reciprocal Identities:
𝑦
1
sin 𝜃 = =
𝑟 csc 𝜃
𝑟
1
csc 𝜃 = =
𝑦 sin 𝜃
Quotient Identities:
𝑥
1
cos 𝜃 = =
𝑟 sec 𝜃
𝑟
1
sec 𝜃 = =
𝑥 cos 𝜃
tan 𝜃 =
Pythagorean Identities:
sin2 𝜃 + cos 2 𝜃 = 1
𝑦
𝑥
=
sin 𝜃
cos 𝜃
𝑦
1
tan 𝜃 = =
𝑥 cot 𝜃
𝑥
1
cot 𝜃 = =
𝑦 tan 𝜃
cot 𝜃 =
1 + tan2 𝜃 = sec 2 𝜃
𝑥
𝑦
=
cos 𝜃
sin 𝜃
1 + cot 2 𝜃 = csc 2 𝜃
These come from the fact that in a right triangle with sides x,y, and r we have: 𝑥 2 + 𝑦 2 = 𝑟 2
2
2
2
2
2
𝑥
𝑦
𝑥
+
𝑦
𝑟
sin2 𝜃 + cos 2 𝜃 = 2 + 2 =
= 2=1
2
𝑟
𝑟
𝑟
𝑟
Trigonometric Identities can be very useful
when trying to do the following problem.
If 𝜃 is an angle in the second quadrant and
5
sin 𝜃 = find the other trigonometric
6
functions of t.
2
5
2

cos
t 1
 
6
25
 cos 2 t  1
36
11
2
cos t 
36
11 Second quadrant
cos t  
cosine is negative
6
5
6
 11
cos t 
6
sin t 
tan t 
cot t 
sec t 
csc t 
5
sin t
5
 5 11
 6 

cos t  11  11
11
6
1
 11

tan t
5
1
1
 6 11


cos t  11
11
6
1
6

sin t 5
Besides using Identities you can equate
parts of fractions.
𝑦 − 3
sin 𝜃 = =
𝑟
2
𝑥 −1
If sec 𝜃 = −2 and the value of 𝜃 is in the
cos 𝜃 = =
𝑟
2
third quadrant, find the value of the other
six trigonometric functions.
𝑦 − 3
tan 𝜃 = =
= 3
𝑥
−1
𝑟
2
𝑥
−1
1
3
sec 𝜃 = =
𝑥 −1
cot 𝜃 = =
=
=
𝑦 − 3
3
3
𝑟 = 2 𝑎𝑛𝑑 𝑥 = −1
𝑟
2
𝑥2 + 𝑦2 = 𝑟2
sec
𝜃
=
=
=
−2
𝑥 −1
1 + 𝑦2 = 4
𝑟
2
2 3 −2 3
𝑦2 = 3
In the third
csc 𝜃 = =
=
=
quadrant y is
𝑦 − 3
−3
3
𝑦=− 3
negative.