Download MR12 Lsn 62

Aim: How do we find the values of trig
functions of quadrantal angles?
Do Now: Given a unit circle and triangle AOB:
M at h Com poser 1. 1. 5
ht t p: / / www. m at hcom poser . com
Find the values of
a) sin 
1
b) cos

c) tan 
HW:
p.380 #
20,22,24,26,28,30,32
p.395 #
22,42,56(a,b,c)
A (0.8, 0.6)
 
-1
O
-1

1
Remember: in unit circle
y
tan  
x
counterclockwise,
x
y
sin    y, cos    x
r
r
If we rotate the terminal side OA
the value of y will keep increasing.
Finally, the terminal side will coincide with the
y-axis and at this moment x = 0 and y = 1.
The   90 Therefore, we say that Sin90  1
If we keep rotating the terminal side, it will
coincide with the negative x-axis at (-1,0).
Therefore, we know that Sin180  0
Sin270  1 and Sin360 / 0  0
.
.
In the same fashion, we can get
cos 90  0, cos180  1 , cos 270  0 and
cos 360 / 0  1
0
tan 0   0
1
1
tan 90  (undefined)
0
0
tan 180 
 0,
1
1
tan 270 
(undefined)
0
Quadrantal Angles
Try This
90
• Find the six

trigonometric
for
 270values
. 1. Draw the angle.
Step
Step 2. Find the ordered pair
from the unit circle..
Step 3. Apply the definitionswe
learned from the reference angle
to find the trigonometric values.
( 0 ,1 )
180
( 1 , 0 )
sin  270  
y 1

r 1 1
x 0

cos 270     0
r 1
y
tan  270    1  undefined
x 0


(1, 0 )
360
1

270
( 0,  1 )
In the unit circle, the radius = 1 that means
the hypotenuse of any right triangle is also 1
1

opp
sin  
 opp.
1
adj
cos  
 adj.
1
opp sin 
tan  

adj cos 
adj cos 
cot  

opp sin 
They are called
quotient identity
How do we use quotient identity to solve
problems?
  2 2 1 
P  3 , 3  on


A point
and cot 
M at h Com poser 1. 1. 5
ht t p: / / www. m at hcom poser . com
y
1
x
θ
-1
P
1
-1
the unit circle, find
tan 
1
sin 
tan  
 3
cos   2 2
1
2
3


4
2 2
1
4
4 2
cot  


2 2
tan 
2
2 2
A circle with center at (0, 0) and radius 1 is
called a unit circle.
(0,1)
(-1,0)
(1,0)
(0,-1)
The points on the circle must satisfy x2 + y2 = 1
For angles that are multiples of 30 they only have one of 2 possible
values for x and y if they don’t land on an axis. If the x is further
you know that one is the square root of 3 over 2, if it is not as far
over as the y is up or down you know it is ½.
 1 3
 , 
 2 2 


120°
 3 1
 , 
 2 2


 1,0
0,1
90°
1 3
 , 
2 2 


60°
150°
30°
 3 1
 , 
 2 2


0° 1,0
180°
210°
 3 1


 2 , 2 


330°
240°
 1
3
  , 
 2 2 


270°
300°
0,1
 3 1
 , 
 2 2


1
3
 , 
2 2 


1. sin 630
2. cos( 630)
3. tan( 900)
4. cos1260
5. tan 720
6. (sin 0)(tan 30)  cos 0
7. sin 45 cos 45  sin 90 cos 90
8. tan( 120) sin 30  cos 30 tan 0
9. cos 405 sin( 45)  tan 135 cos( 30)