Download Unit Circle & Angle Measures

Aim: What good is the Unit Circle and how
does it help us to understand the
Trigonometric Functions?
Do Now:
A circle has a radius of 3 cm. Find the
length of an arc cut off by a central angle of
2700.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Angles in Standard Position
y
Q II
Quadrant I
90 <  < 180

 
0 <  < 90

0 
2

2
initial side
x
Q III
Q IV
180 <  < 270
3
  
270 <  < 360
3
   2
2
2
• An angle on the coordinate plane is in standard
position when its vertex is at the origin and its initial
side coincides with the nonnegative ray of the x-axis.
• An angle formed by a counterclockwise rotation
has a positive measure.
• Angles whose
side
of 2the
Aim:terminal
Trig. Ratios for any
Angle lies on one
Course: Alg.
& Trig.axes is
a quadrantal angle. i.e. 900, 1800, 2700, 3600, 4500 etc.
Co-terminal and Negative Angles
y
Q II
Quadrant I
90 <  < 180

 
0 <  < 90

0 
2
3000 = 
2
-600 initial side
x
Q III
Q IV
180 <  < 270
3
  
270 <  < 360
3
   2
2
2
• An angle formed by a clockwise rotation has a
negative measure
• Angles in standard position having the same
terminal side are co-terminal angles.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Q II
90 <  < 180

 
2
Angles Greater than 3600
y
0
Quadrant I
485
0 <  < 90

0 
2
1250
Q III
Q IVx
180 <  < 270
3
  
270 <  < 360
3
   2
2
2
• Angles whose terminal side rotates more than one
revolution form angles with measures greater
than 3600.
• To find angles co-terminal with an another angle
add or subtract 3600. 1250 and 4850 are co-terminal
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
Find the measure of an angle between 00
and 3600 co-terminal with
a) 3850 25o
b) 5750 215o c) -4050 315o
In which quadrant or on which axis, does
the terminal side of each angle lie?
x0
0
a) 150 QII b) 540
c) -600 QIV
axis
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
radius = 1
center at (0,0)
Unit Circle
1 y
cos , sin 
(x,y)

-1
cos 
side adj.
lengthside adj . to 
cos 
length hypo.
1
x
length side opposite 
sin 
length  hypo.
length side opposite 
lengthside adj . to 
sin 
cos 
-1
1
1
cos  lengthside adj . to 
sin  length side opposite 
cos  x
Aim: Trig. Ratios for any Angle
sin  y
Course: Alg. 2 & Trig.
Aim: What good is the Unit Circle and how
does it help us to understand the
Trigonometric Functions?
Do Now:
Find the measure of an angle between
00 and 3600 co-terminal with an angle
whose measure is -1250.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Value of Sine & Cosine: Quadrant I
y
1
radius = 1
cos 600, sin 600
(x,y)
center at (0,0)
1 3 
 ,

2 2 
600
x
1
What is the value of coordinates (x,y)?
-1
300-600-900 triangle
Hypotenuse = 2  shorter leg
Longer leg =
3  shorter leg
Sine and Cosine values for angles in
Quadrant I are positive.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Value of Sine & Cosine: Quadrant II
y
1
0
0
cos 120 , sin(x,y)
120
What is the value
 1 3 
 ,

of coordinates (x,y)?
 2 2 
1
60º is the
600 1200
reference
x
-1
angle
1
(180º-120º)
A
reference angle
for any angle in
standard
position is an acute
angle formed by the
terminal
side of the given
angle and the
x-axis.
directed
sidedistance
adj.
What is the cosine/sine of a 1200 angle?
300-600-900 triangle
Hypotenuse = 2  shorter leg
Longer leg =
3  shorter leg
Sine values for angles in Quadrant II are positive.
Cosine values for angles in Quadrant II are negative.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Value of Sine & Cosine: Quadrant III
1 y
What is the value
of coordinates (x,y)?
(240º-180º)
2400
directed
side distance
adj.
directed dist.
60º is the
reference
-1
angle
What is the cosine/sine
of a 2400 angle?
600
1
x
1
0
(x,y)
cos 2400, sin 240
 1
 , 3 
 Sine and Cosine values for angles in
 2
2  Quadrant III are both negative.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Value of Sine & Cosine: Quadrant IV
1 y
What is the value
of coordinates (x,y)?
3000
600
(360º-300º)
1
directed dist.
60º is the
reference
-1
angle
What is the cosine/sine
of a 3000 angle?
1
x
1
 , 3 

2
2 
(x,y)
cos 3000, sin 3000
Sine values for angles in Quadrant IV are negative.
Cosine values for angles in Quadrant IV are positive.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Unit Circle – 12 Equal Arcs
1.2
5
6
,
5
6
-
1
2
,
3
1
1
2
+ 2. . .
2
,
3
2
6
0.8
3 1
,
2 2

,

6
+ 2. . .
3 1
,
2 2
0.6
0.4
0.2
-1
-0.5
=

6
0.5
1
-0.2
-0.4
7
6
,
7
6
3
2
,-
1
3
-0.6
2
-0.8
+ 2. . .
-
1
2
,-
sin  cos(
3
-1
2

2
-1.2
 )
Aim: Trig. Ratios for any Angle
1
2
,-
3
,-
1
2
2
11 11
,
+ 2. . .
6
6
2
cos   sin(

2
 )
Course: Alg. 2 & Trig.
Unit Circle – 8 Equal Arcs
1.2
3
4
,
3
4
-
+ 2. . .
2
,
2
1
2
Periodic
4
2
0.8
2

2
0.6

,
4
+ 2. . .
2
,
2
0.4
0.2
-1
-0.5
=

4
0.5
1
-0.2
-0.4
-0.6
-
2
,-
2
-0.8
2
,-
2
2
Negative Angles
7 7
,
Identities
+ 2. . .
4
4
sin(  )   sin 
 )  cosCourse: Alg. 2 & Trig.
Aim: Trig. Ratioscos(
for any Angle
2
2
2
-1
5
4
,
5
4
-1.2
+ 2. . .
Value of Sine & Cosine in Coordinate Plane
y
Quadrant II
Quadrant I
cos  is –
sin  is +
cos  is +
sin  is +
Quadrant III
Quadrant IV
cos  is –
sin  is –
cos  is +
sin  is –
x
The reference angle: for any angle in standard
position is an acute angle formed by the terminal
side of the given angle and the x-axis.
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
Fill in the table
Quad.
Ref. 
sin
cos 
a) 2360
b) 870
c) -1600
d) -36
e) 13320
f) -3960
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Regents Prep
On the unit circle shown in the diagram
below, sketch an angle, in standard position,
whose degree measure is 240 and find the
exact value of sin 240o.
1
y
-1
1
x
-1
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Aim: What good is the Unit Circle and how
does it help us to understand the
Trigonometric Functions?
Do Now:
Use the unit circle to find:
a. sin 1800 () b. cos 1800
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
Use the unit circle to find:
a. sin 1800 () b. cos 1800

(-1,0)
(x, y) = (-1, 0)
sin 1800 = y = 0
cos 1800 = x = -1
180º - quadrantal angle
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Tan 
( 1,, tan)
)?
radius = 1
center at (0,0)
1 y
cos , sin 
sin 
tan 
cos 
(x,y)
y
  y
1
1
-1

cos
cos  = 1
1
x
length of the leg opposite 
tan 
length of the leg adjacent to 
sin
y
-1
tan  

cos x
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Trigonometric Values
Quad.
Sin 
Cos 
I
+
+
+
–
II
Sin  =
reference
angle Sin(180– )
III
–
reference
Sin  =
angle -Sin( -180)
IV
–
Sin  =
reference
angle -Sin(360- )
Cos  =
-Cos(180– )
–
Cos  =
-Cos( -180)
+
Cos  =
Cos(360-)
Aim: Trig. Ratios for any Angle
Sin
Cos



–
–
–
–

Tan 
+
–
Tan  =
-Tan(180– )
+
Tan  =
Tan( -180)
–
Tan  =
-Tan(360- )
Course: Alg. 2 & Trig.
Trigonometric Values - A C T S
y
Q II
Quadrant I
90 <  < 180

 
2
0 <  < 90

0 
2
S
A
Sine is +
All are +
T
x
C
Tangent is + Cosine is +
Q III
Q IV
180 <  < 270
3
  
2
270 <  < 360
3
   2
2
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Need to Knows
1y
When r = 1
sin  = y
cos  = x
r
-1
tan  = y/x
Reciprocal Functions
csc  = 1/y
sec  = 1/x
cot  = x/y

x
y
1 x
-1
Negative Angles
Identities
sin(  )   sin 
cos(  )  cos
tan(  )   tan
denominators  0
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
y
Using the unit circle, find 1
(x,y)
450 (/4)
a. cos
b. sin 450
c. tan 450
1
-1
450
450-450-900 triangle
cos  = x
In a 450-450-900 triangle, the length of the
hypotenuse is 2 times the length
-1of a leg.
1
A 450-450-900 triangle is an isosceles right triangle.
therefore x = y
cos  = sin 
length of hypo. = 2 (x)
1  2(x)
1
 1 
2 
2


x

x y
 2  2  2
2
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
x
Model Problems
y
1
Using the unit
 2 2


,
(x,y)

 2 2 
circle, find
1
a. cos 45º(/4)
b. sin 45º
-1
c. tan 45º
2
xy
2
2
cos 45º = x =
2
2
sin 45º = y =
2
45º
cos  = x
1
-1
2
sin45 y
tan45 
  2
cos 45 x
2
2
tan 45 = 1
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
x
Trigonometric Values for Special Angles
0º
0

sin 
0
cos 
1
tan 
0
30º
/6
1
2
3
2
3
3
45º
/4
60º
/3
90º
/2
2
2
2
2
3
2
1
2
1
1
3
UND.
0
Why is tan 90º undefined?
What is the slope of a line
perpendicular to the x-axis?
sin
y

cos  x sin  y tan  
= slope
cos x
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
What is the tan 135º (3/4)?
• 135º is in the 2nd quadrant
• 45º is reference angle (180 – 135 = 45)
• tan 45º = 1
• tangent is negative in 2nd quadrant
• tan 135º = -1
What is the cos 510º (17/6)?
• 510º is in the 2nd quadrant
(510 – 360 = 150)
• 30º is reference angle (180 – 150 = 30)
3
2
• cos 30º =
• cosine is negative in 2nd quadrant
3
• cos 510º = 
≈ -.866…
Aim: Trig. Ratios for any2
Angle
Course: Alg. 2 & Trig.
Model Problems
1
3
Point A ( , 
) is on the unit circle whose
2
2
center is the origin. If  is an angle in
standard position whose terminal ray passes
through point A. Find the value of of the six
trig. functions.
a) cos
b) sin
c) tan
d) sec
e) csc
f) cot
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problems
Given:
sin 68o = 0.9272
cos 68o = 0.3746
Find cot 112o
sin
tan 
cos
A) -0.3746
B) -2.4751
C) -0.404
D) 1.0785
reference angle for 112o is 68o;
WHAT
112o is
in QII;ELSE DO WE KNOW?
tan and cot are negative in QII
1
1
cos112
o
cot112 


sin112
tan112
sin112
cos112
cos68
0.3746


 0.404
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
sin 68
0.9272
Model Problems
Express sin 285º as the function of an angle
whose measure is less than 45º.
What do we know?
285º in IV quadrant
the sine of a IV quadrant
angle is negative
reference angle for 285º is
(360 – 285) = 75º
-sin 75º
> 45º
sine and cosine are co-functions
complement of 75º is 15º
sin 285º = -sin 75º = -cos 15º < 45º
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Trig Functions Using Radian Measures
Algebraically:
Find:
sin (π/3)
remember:
π/3
π/3 radians
sin 60º =
mA 0
180 0

mA in radians

60º
3
≈ .866…
2
Using the calculator:
Use the mode key:
change setting from degrees to radians
then hit:
sin 2nd
π
÷ 3 ENTER
Display: .8660254083
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Un-unit circle
 is any angle in
r1
standard position
with (x, y) any point
on the terminal side
of  and r  x 2  y 2
1
-1
y
1
1 x
unit-1circle
cos  x
sin  y
y
sin  
r
x
cos  
r
y
tan  
x
Aim: Trig. Ratios for any Angle
r
csc  , y  0
y
r
sec  , x  0
x
x
cot   , y  0
y
Course: Alg. 2 & Trig.
Model Problem
(-3, 4) is a point on the terminal side of .
Find the sine, cosine, and tangent of .
r
x y
2
(-3, 4)
2
r  ( 3 ) 2  4 2
r  25  5
y 4
sin   
r 5
x 3
3
cos   

r
5
5
y
4
4
tan   

x 3
3
Aim: Trig. Ratios for any Angle
4
-5
4
2
r= 5

3
1 4
sin
 53.130
5
-2
-4
Q II
  180  53.130
 126.897
Course: Alg. 2 & Trig.
Model Problem


r
x2  y2
3, 1 is a point on the terminal side of .
Find , the sine, cosine, and tangent of .
r  (  3)   1
2
r 42
y
1
sin 

r
2
x
 3
cos 

r
2
y
tan   1
x  3
2
 3
-1

3, 1

r=2
1
sin    300 Q III
 2
7
  210 or
radians
Aim: Trig. Ratios for any Angle
1
6
Course: Alg. 2 & Trig.
Model Problem
Tan  = -5/4 and cos  > 0, find sin  and sec 
When tangent is negative
and cosine is positive
angle is found in Q IV.
4
Quadrant II
sin  : +
cos  : tan  : -
2
Quadrant I
sin  : +
cos  : +
tan  : +
5
y 5
tan   
x
4
r
Quadrant III
sin  : cos  : tan  : +
-2
Quadrant IV
sin  : cos  : +
tan  : -
-4
x 2  y 2  4 2  ( 5)2  41
5
y
 0.7809
sin  
r
41
r
41
sec 

 1.6008
x
4
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Model Problem
The terminal side of  is in quadrant I and
lies on the line y = 6x. Find tan ; find .
y = mx + b - slope intercept form of equation
y
 tan
m = slope of line 
x
y = 6x
m = 6 = tan 
tan 1 6  80.538
Aim: Trig. Ratios for any Angle
QI
Course: Alg. 2 & Trig.
Model Problem
The terminal side of  is in quadrant IV and
lies on the line 2x + 5y = 0. Find cos .
2
y = mx + b
y   x slope intercept
5
form of equation
tan  = m = -2/5
y sin 
2
tan  
   x  5; y  2
x cos 
5
r x y 
2
2
 2    5 
2
2
 29
x
5
5 29
cos  

r
29
29
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.
Templates
1
y
1
45º
-1
1
x
-1
Aim: Trig. Ratios for any Angle
Course: Alg. 2 & Trig.