Download Trig Identities Practice

Evaluating Trig Functions
I.
Definition and Properties of the Unit Circle
a. Definition: A Unit Circle is the circle with a radius of one ( r  1 ),
centered at the origin (0,0) .
b. Equation: x 2  y 2  1
c. Arc Length
Since arc length can be found using the formula: s  r
(where s = arc length, r = radius,  = central angle in radians)
For the unit circle, since r = 1,
s  (1)
Therefore s  
The arc length of a sector of a unit circle
equals the radian measure of angle  .
d. Circumference: C  2r  2 (1)  2
The arc length (circumference) of 2 is also the radian measure
of the angle corresponding to 360 .
2 radians = 360 degrees
 radians = 180 degrees
e. Relating Coordinate Values to Trig Functions
For any point P( x, y) on the unit circle,
x  cos  and y  sin  where  is
any central angle with:
1) initial side = positive x axis
2) terminal side = radius through pt. P
In the first quadrant this can be verified:
opp y y
sin  
  y
hyp r 1
cos  
adj x x
  x
hyp r 1
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f. The x and y axes can be labeled using the radian measure of the
angle  which corresponds to the points where the unit circle
intersects the axes.
(0,1)
(1,0)
(-1,0)
0, 2
(0,-1)
We can also see this graphically:
y  sin x
( ,1)
2
(2 ,0)
( ,0)
(0,0)
(3 ,1)
2
y  cos x
(2 ,1)
(0,1)
( ,0)
2
(3 ,0)
2
( ,1)
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y  tan x
(0,0)
( ,0)
(2 ,0)
Finding the sine and cosine values of quadrantal angles is now easy.
3
For example, to find sin
use the point (0,-1) which corresponds
2
3
3
to a central angle of
. Since sin
is the y coordinate of the
2
2
3
3
point, sin
 1 . Similarly, cos
 0 (the x coordinate of the
2
2
point).
g. Examples:
i. Find the arc length of a sector in a unit circle with a central
angle of 120 .
Solution:
In a unit circle, arc length = central angle measured in radians,
s   . Since  radians equals 180 , multiply 120 by the conversion
 radians
ratio of
.
180
   2
Thus the arc length and the measure of  are
120 
=
 180  3
2
both
radians.
3
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ii. Find cos

2
and sin

2
.
Solution:

cos
P).

2

2
implies the angle is a right angle ( 90 ), so P  (0,1) . Hence,
 0 (the x coordinate of P) and sin

2
 1 (the y coordinate of
iii. Find sin    and cos  
Solution:
If    , P  (1,0) . So sin    0 (y value) and cos    1 (x
coordinate of P).
II.
More Properties of the Unit Circle
a. If  

4
(which is equivalent to 45 ), then for the point P on the
unit circle, x  y 
2
.
2
P( 2
If  
 2 2
.
( 45 ), then P  
,

4
2
2



1

x
2
, 2 )
2
y
Explanation:
x 2  y 2  1 (equation of unit circle)
x 2  x 2  1 ( x  y since it is an isosceles 45  45  90 triangle)
2x 2  1
1
1
2
x


y
2
2
2
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b. If  

6
(which is equivalent to 30 ), then for the point P on the
1
3
and y  .
2
2
unit circle, x 
Using properties of 30  60  90 triangles with hypotenuse of
length 1 (since r  1 ):
y
If  
 3 1
( 30 ), then P  
,  .
2
2
6


c. If  

3
1

 3 1
P  
, 
 2 2
1
2
3
2
x
(which is equivalent to 60 ), then for the point P on the
unit circle, x 
1
3
and y 
.
2
2
Using properties of 30  60  90 triangles with hypotenuse of
length 1 (since r  1 ):
y
1 3

P   ,

2 2 
1

3
2
1
2
If  
x
1 3
.
( 60 ), then P   ,

2
2
3



NOTE:
The larger side, 3 , is always opposite the larger angle,  ,
3
2
and the smaller side, 1 , is opposite the smaller angle,  .
2
6
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d. Examples:
Find :
(i) sin

(ii) csc
6


(iii) tan
3
3
3
( , )
2 2
(

4

2 2
,
)
2 2

1

(y coordinate when   )
6 2
6

1
1
2
2 3
(ii) csc 




3
3
3
3
sin
3
2

2
 sin 4
(iii) tan 
 2 1

4 cos
2
4
2
(i) sin


6
3 1
( , )
2 2
x
III.
4
Solution:

y1


More on Evaluating Trig Functions Using the Unit Circle
It is important to recognize the radian measure of the standard angles

 
, , and
located in quadrants II, III, and IV.
6 4
3
related to
Graph
y
x
y
x
y
x
Reference
Angle
QI
QII
QIII
QIV
Point P

6

6
5
6
7
6
11
6

3 1


 2 , 2 



4

4
3
4
5
4
7
4

2
2


 2 , 2 



3

3
2
3
4
3
5
3
 1
3
  ,

 2
2 

The signs of the x and y values of point P can be determined by knowing
the quadrant the angle terminates in. It’s as simple as remembering:
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All Students Take Calculus
Quadrant I
Quadrant II
Sine & cosecant
are positive
are positive
Quadrant III
Quadrant IV
Cosine & secant
Tangent &
cotangent are
positive
Examples:
1.
All trig functions
Evaluate sin
are positive
5
.
4
Solution:
5

has a reference angle of .
4
4
5

   , it is in quadrant III.
Since
4
4
P  (
sin
2
2
,
)
2
2
5
2

4
2
P
NOTE: Our final answer will be
negative because only tan and cot 
are positive in the third quadrant.
(since x and y are negative in QIII)
(the y coordinate of P)
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2.
Evaluate cos
5
.
6
P
Solution:
5

has a reference angle of .
6
6
Since
P  (
cos
3.
5

   , it is in quadrant II.
6
6
3 1
, )
2 2
NOTE: Our final answer will be negative
because only sin  and csc are positive
in the second quadrant.
(since x negative in QII)
5
3

6
2
(the x coordinate of P)
Evaluate tan
11
.
3
Solution:
11

has a reference angle of .
3
3
Since
11

 2  , it is in quadrant IV.
3
3
P
NOTE: Our final answer will be negative
because only cos  and sec are positive
in the fourth quadrant.
1
3
P  ( ,
)
2
2

3
11
3  2  3
tan


1
3
cos
3
2
sin
IV.
Other Facts Derived From the Unit Circle
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Page 8 of 10
1.
Fundamental Identity:
For any point P on the unit circle, P  ( x, y)  (cos  , sin  ) .
Substituting x  cos  and y  sin  into the equation of the circle:
x2  y 2  1
(cos  )2  (sin  )2  1
sin 2   cos 2   1
2.
Other identities:
The x coordinates of points for  and   are the same, so:
cos( )  cos( )
Therefore cos( ) is an even function.
The y coordinates of points for  and   are the opposite, so:
sin( )   sin( )
Therefore sin( ) is an odd function.
( x, y)


( x, y)
Practice Exercises:
Use the unit circle to answer the
following problems.
1. Evaluate:
a. cos 
b. sin
3
2
2. Find the six trigonometric
functions values at the
following values of  :
5
a.
3
c. tan(3 )
d. csc

b.
3
4
c.
7
6
d. 
2
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11
6
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Solutions: Unit Circle Trig
1. a) -1
b) -1
c) 0
d) 1
2.
a.
Angle
Quadrant
Point P
5
3
7
6
c.
d. 
11
6
QII
QIII
QI
1
3
 ,

2

2



2 2


 2 , 2 



3 1


 2 , 2 


 3 1


 2 ,2



3
2
cos 
1
2
tan
 3

2
3
sec
cot 
3
4
QIV
sin 
csc
b.

2
2

1
3
2
2

1
1


3
3
3
2
 2
2
2 3
3
2


2
 2
2
1
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1
2
1
2
3
2
3
2

3
3
1
3
2

2
3

3
3
3

2
2 3
3
2
3

2 3
3
3
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