Download Lecture 4B: Trigonometric Functions II (WORD)

Other Trigonometry Functions
Tangent:
tan x 
sin x
cos x
2 n  1
Undefined if cos x  0 , or x  

2


Cotangent:
cot x 
cos x
sin x
Undefined if sin x  0 , or x  n 
Secant:
sec x 
1
cos x
 2 n  1
Like Tangent, undefined for x  

2


Cosecant:
csc x 
1
sin x
Like Cotangent, undefined for x  n 
In all the above expressions, n  0,  1,  2, 
In terms of the sides of a right triangle,
tan x 
OPP
ADJ
MATH 1310
Ronald Brent © 2016 All rights reserved.
cot x 
ADJ
OPP
sec x 
Lecture 4B
HYP
ADJ
csc x 
HYP
OPP
1 of 12
Graphs of Other Trigonometry Functions
y
y = tan x
5
4
3
2
1
0

5
3
5
3


 
   


2
2
2
2
2
2

x
-1
-2
-3
-4
-5
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
2 of 12
y
y = sec x
5
4
3
2
1
 0
5
3
5
3


 
   


2
2
2
2
2
2

x
-1
-2
-3
-4
-5
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
3 of 12
y
y = cot x
5
4
3
2
1
0

x
-1
-2
-3
-4
-5
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
4 of 12
y = csc x
y
5
4
3
2
1
0

x
-1
-2
-3
-4
-5
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
5 of 12
Solving Right Triangles:
Example: Solve:
32
o
8

We know that the side adjacent to the 32 angle is ADJ = 8.
To get OPP, the side opposite the angle, consider that in this case
OPP
 tan 32 ,
ADJ
and so
OPP  ADJ tan32  (8)  (.624869352)  4.998954815 .
Given the opposite side, OPP, you can get the hypotenuse, HYP, from the Pythagorean theorem, or
by using
HYP
 sec32 ,
ADJ
to obtain
HYP = ADJ  sec 32  (8)  (1.179178403)  9.433427227 .

Of course, the last angle is 58 .
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
6 of 12
Non-Right Angle Triangles
Given the triangle show below with sides of length a, b, and c, and opposite angles A, B, and C,
we have:
a
b
C
B
A
Law of Sines
c
sin A sin B sin C


a
b
c
Law of Cosines
c 2  a 2  b2  2 a b cos C
also
b2  a 2  c 2  2 a c cos B
and
a 2  b2  c 2  2 b c cos A
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
7 of 12
Example: Solve:
60
o
25o
15
Using the Law of Sines
c
A
B = 60
o
b
a = 15
C = 25
o

Notice that A = 95 , so
sin 95 sin 60 sin 25


.
15
b
c
You can solve for b and c as
15 sin 60
b
sin 95
15 sin 25
and c 
.
sin 95
Using a calculator, you can get decimal equivalents:
b  13.04 and c  6.36
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
8 of 12
Example:
Solve
4
75
o
5
First, label the triangle:
A
b
c = 4
B = 75 o
C
a = 5
b2  a 2  c 2  2ac cosB  25  16  40 cos 75
 38.32
 b  6.19 .
Now you can find the angle C from the Law of Sines:
sin B sin C
sin 75 sin C


, so
and
6.19
4
b
c

sin 75
sin C  4 
 0.624
6.19
C = angle, between 0 and 90, whose sine is .624.


Using a calculator we find out that C  38.6 A  66.4 .
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
9 of 12
Trigonometric Identities
y
(cos  , sin  )


x
(1,0)
(cos  , sin  )
sin( )   sin
cos( )  cos


sin   cos   
2

MATH 1310
Ronald Brent © 2016 All rights reserved.
tan( )   tan


cos  sin   
2

Lecture 4B
10 of 12
Given
sin 2   cos 2   1 ,
2
if we divide both sides of this equation by cos  we get
sin 2   cos 2 
1

cos 2 
cos 2 
sin 2  cos 2 
1


cos 2  cos 2  cos 2  ,
tan 2   1  sec 2  .
2
Similarly if we divide by sin  , we can derive
1  cot 2   csc 2  .
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
11 of 12
Sum and Difference Formulas:
sin( A  B )  sin A cos B  sin B cos A
sin( A  B )  sin A cos B  sin B cos A
cos( A  B )  cos A cos B  sin A sin B
cos( A  B )  cos A cos B  sin A sin B
Double Angle Formulas
sin 2  2 sin  cos
cos 2  cos2   sin 2 
Half Angle Formulas
1  cos 
2
cos 2 12  
1  cos ( 2 )
2
cos 2  
sin 2 12  
sin 2  
MATH 1310
Ronald Brent © 2016 All rights reserved.
Lecture 4B
1  cos
2
1  cos ( 2 )
2
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