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SFM Productions Presents:
Another saga in your continuing Pre-Calculus experience!
4.3 Right Triangle Trigonometry
Homework for section 4.3
p306 # 7-35 eoo, 37-45, 49, 51, 57-
61, 67, 70, 72
opposite
The six trigonometric functions:
θ
adjacent
sin  
op p osit e
hyp ot enuse
cos  
adjacent
t an  
csc  
sec  
cot  
hyp ot enuse
op p osit e
sin 

adjacent
cos 
hyp ot enuse
op p osit e
hypot enuse
adjacent
adjacent
opposit e

cos 
sin 
Find the trigonometric functions for the following:
Use the
Pythagorean Theorem
to find the hypotenuse.
5
4
θ
3
sin  
o p p o sit e
hyp o t en use

4
5
co s  
adjacen t

3
5

4
3

5
4

5
3

3
4
hyp o t en use
t an  
o p p o sit e
adjacen t
csc  
hyp o t en use
sec  
hyp ot enuse
cot  
o p p o sit e
adjacent
adjacent
op p osit e
Trigonometric Identities: You should
already have these
1
sin  
csc 
1
cos  
sec 
1
csc  
sin 
1
sec  
cos 
1
t an  
cot 
1
cot  
t an 
sin 
t an  
cos 
cos 
cot  
sin 

1

sin   
6 
2

1

co s   
3 
2

3

sin   
3 
2

3

co s   
6 
2


sin 
    cos 
2



cos 
    sin 
2



t an      cot 
2



cot      t an 
2



sec      csc 
2



csc      sec 
2

These are the: Co-function Identities
Trigonometric Identities
Pythagorean Identities
a 2  b2  c2
c
x
2
 y
2
 r
a
r
2
b
y
x
(This is also the formula for a circle…)
sin   cos   1
2
B y def init ion:
2
sin 2  is t he same t hing as
( and similar not at ion f or t he ot her t rig f unct ions)
sin  2
sin   cos   1
2
2
sin12 θ
sin2 θ
cot22θθ
+ cos
sin2 θ
= csc21θ
sin2 θ
sin22θθ
tan
cos2 θ
+ cos12 θ
cos2 θ
= sec12 θ
cos2 θ
Two more identities can be derived from this
identity.
Example:
Let θ be an acute angle such that
sin θ = 0.6
Find the values of a) cos θ b) tan θ
op p
0 .6

sin   0 .6 
hyp
1
1
0.6
sin 2   cos 2   1
2
2
0
.6

cos
  1
 
cos   1  0 .6 
cos 2   0 .6 4
cos   0 .8
2
θ
0.8
2
sin 
0 .6
t an  

 0 .75
cos 
0 .8
Example:
A surveyor is standing 50 feet from the base of a large tree.
The surveyor measures the angle of elevation to the top of the
tree as 71.5◦. How tall is the tree?
t an 71.5
t an 7 1.5
op p

adj

y
50
y  5 0 t an 71.5
y
71.5◦
50 ft
y  149 .4 f eet
Example:
300
You are on a patch of grass 300 feet distant from a lake. You decide to walk in a straight line at
an angle towards the lake. The distance you travel is 500 feet. What is the angle made between
your path and the edge of the lake?
θ
500
op p
300
sin  

hyp
5 00
sin   0 .6
  sin 1 0 .6 
  3 6 .8 7
sin   0 .6
value of the angle
angle
  sin
angle
1
0 .6 
value of the angle
 
sin    y
6 
Given the angle;
1
sin  
2
Given the value of the trig function
of an angle;
Find the value of the trig function of
that angle.
Find the angle.
  sin
1
 1
 
2 
Go! Do!
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