Algebra III
Lesson 36
Angles Greater than 360° - Sums of
Trigonometric Functions – Boat-in-theRiver Problems
Angles greater than 360°
Whether the angle is 730° or -810°, the only trick is to remove
multiples of 360° until the angle is less than 360°.
The problem is then handled like any other angle problem
from the past.
Sums of trigonometric functions
There is nothing new here.
Using the unit circle will make these problems go faster.
The unit circle will never be provided. If you prefer
these can be done using reference triangles.
Answers are to be combined until there is only one term,
combine everything.
Example 36.1
Evaluate: cos 135° + tan 330°
cos 135° + tan 330°
2
3
=−
−
2
3
(-1,0)
0°
θ=6
0°
θ=3
θ=180°

2
2
−

,
−
 2

2


θ=0°
θ=360°
10°
2
=
θ
θ=3
30°
5°
2
2
θ=
 1
3
 − ,−

 2

2


θ=
3
(0,-1)
15
°
00°
−3 2 −2 3
6
 2 2


 2 , 2 


5°
4
=
θ=1
50°

3 1
−

 2 ,− 2 


=
θ
θ=3
3 2 2 3
−
6
6

3 1

−
 2 , 2


θ=270°
=−
θ=
13
5°
20°
1
−1
y
3
tan 330° = = 2 =
=−
3
3
x
3
2
−
1 3
 ,

2 2 


(0,1)
θ=90°

2 2

−
 2 , 2 


θ=1
2
2
θ=
24
0°
cos135° = −
 1 3

− ,
 2 2 


1
3
 ,−

2

2


 3 1


 2 , 2


(1,0)
 3 1


 2 ,− 2 


 2
2


,
−
 2

2


Example 36.2
Evaluate: cos (-60°) + cos 210°
cos (-60°) + cos 210°
1
3
= −
2 2
(-1,0)
0°
θ=6
θ=90°
θ
0°
θ=3
θ=180°

3 1
−

 2 ,− 2 


θ=0°
θ=360°
10°
2
=
θ
θ=3
30°
5°
2
2
θ=
 1
3
 − ,−

 2

2


θ=
3
(0,-1)
15
°
00°

2
2
−

,
−
 2

2


 2 2


 2 , 2 


5°
4
=
θ=1
50°
θ=3
1− 3
2

3 1

−
 2 , 2


θ=270°
=
θ=
13
5°
1 3
 ,

2 2 


(0,1)
20°
3
2

2 2

−
 2 , 2 


θ=1
cos 210° = −
1
2
θ=
24
0°
cos(−60°) = cos 300° =
 1 3

− ,
 2 2 


1
3
 ,−

2

2


 3 1


 2 , 2


(1,0)
 3 1


 2 ,− 2 


 2
2


,
−
 2

2


Example 36.3
Evaluate: cos 570° + sin (-765°)
sin( −765°) = sin( −45°) = −
2
2

3 1

−
 2 , 2


(-1,0)
20°
765 – 720 = 45
θ=
13
5°
0°
θ=6
3
cos 570 = cos 210° = −
2
1 3
 ,

2 2 


(0,1)
θ=1

2 2

−
 2 , 2 


 1 3

− ,
 2 2 


θ=90°
570 – 360 = 210
θ
5°
4
=
0°
θ=3
θ=1
50°
θ=180°
θ=0°
θ=360°
cos 570° + sin (-765°)
 1
3
 − ,−

 2

2


θ=
3
(0,-1)
15
°
00°

2
2
−

,
−
 2

2


5°
2
2
θ=
θ=3
− 3− 2
2
θ=3
30°
θ=270°
=

3 1
−

 2 ,− 2 


10°
2
=
θ
θ=
24
0°
3
2
=−
−
2
2
 2 2


 2 , 2 


1
3
 ,−

2

2


 3 1


 2 , 2


(1,0)
 3 1


 2 ,− 2 


 2
2


,
−
 2

2


Boat-in-the-water problems
These problems all have to do with a vehicle moving on or through a
moving medium.
Mostly these problems, at least at first, will deal with perfectly with
or against the moving medium.
Two basic situations:
1) A boat going up or down a current of water (river)
2) A plane flying with a tailwind or headwind.
Treat these problems as rate-time-distance problems.
Down current or a tailwind
In these situations the speeds combine to make the vehicle go faster.
R·T=D
becomes
(Rvehicle + Rmedium)·T=D
Up current or a headwind
In these situations the speeds combine to make the vehicle go slower.
R·T=D
becomes
(Rvehicle - Rmedium)·T=D
Example 36.4
One day Selby found that her plane could fly at 5 times the speed of
the wind. She flew 396 miles downwind in ½ hour more than it took
her to fly 132 miles upwind. What was the speed of her plane in still
air and what was the speed of the wind?
This problem has two situations upwind and downwind, set each
up separately.
Downwind:
Upwind:
(Rvehicle + Rmedium)·T=D
(5Rw + Rw)·(T + ½)=396
(6Rw)·(T+ ½ )=396
6RwT + 3Rw = 396
(Rvehicle - Rmedium)·T=D
(5Rw - Rw)·(T )=132
4RwT=132
RwT=33
6RwT + 3Rw = 396
Rw = 66 mph
6(33) + 3Rw = 396
Rp = 5Rw = 330 mph
Rw = (396 – 198)/3
Practice
a) Bruce found that his plane could only fly at 4 times the speed of
the wind. He flew 800 miles downwind in 1 hour more than it took
him to fly 300 miles upwind. What was the speed of the plane in still
air and what was the speed of the wind?
(4W – W)·T = 300
(4W + W)·(T + 1) = 800
3WT = 300
5WT + 5W = 800
WT = 100
500 + 5W = 800
5W = 300
W = 60 mph
P = 240 mph
b) Evaluate. Do not use a calculator.
1) cos 225° + tan (-135°)
2
+1
2

3 1

−
 2 , 2


2) sin (-390°) + cos 495°
(-1,0)
0°
θ=90°
θ=6
θ=
13
5°
1 3
 ,

2 2 


(0,1)
20°
=−

2 2

−
 2 , 2 


 1 3

− ,
 2 2 


θ=1
2
−
2
=−
+ 2
2
2
−
2
θ
5°
4
=
0°
θ=3
θ=1
50°
θ=180°
θ=0°
θ=360°
= sin (-30°) + cos 135°
1+ 2
2

2
2
−

,
−
 2

2


 1
3
 − ,−

 2

2


θ=
3
(0,-1)
15
°
00°
=−
5°
2
2
θ=
θ=3
− 1− 2
2
θ=3
30°
θ=270°
=

3 1
−

 2 ,− 2 


10°
2
=
θ
θ=
24
0°
1 
2


= − +−
2  2 
 2 2


 2 , 2 


1
3
 ,−

2

2


 3 1


 2 , 2


(1,0)
 3 1


 2 ,− 2 


 2
2


,
−
 2

2


c) Evaluate:
arccos
3

sin  arccos 
7

3
7
adj
cosθ =
hyp
not on unit circle so use triangle
7
adj=3; hyp=7
θ
opp
sin θ =
hyp
3
need opposite
(opp)2 + 32 = 72
opp = 40
3

sin  arccos  = 40 = 2 10
7

7
7
d) Jan traveled m miles at p miles per hour and arrived two hours early.
How fast should she have traveled in order to arrive on time?
d= m
r= p
t= d/r =m/p
d= m
r= ?
t=
r=
d
t
=
m
p
•
m
p
+2
p
=
mp
m+2p
m
+2
p