Download Radian Measure: Arc length

Feb 12: Discussion of Trig for Calculus
"Trigonometry " is from Greek for "Triangle " + "Measure"
Labeling Convention: Upper Case denotes angles opposite edge denoted
by lower case
Angles are measured by a portion of the circle centered at their vertex
which they subtend
“Straight Angle = ½ circle”
Right angle = ¼ circle.
Babylonians (probably) divided the circle into 360 equal pieces
( degrees) because 360 has lots of divisors:
Fractional parts of circle would have whole numbers of
degrees for lots of fractions.
1/2 circle = 180 degrees
1/3 circle = 120 degrees
1/4 circle = 90 degrees
1/5 circle = 72 degrees
1/6 circle = 60 degrees
1/7 circle ( not an integer number of degrees- it doesn’t always work)
1/8 circle = 45 degrees
Etc.
Understanding of parallel lines from plane geometry (alternate interior
angles) gives sum of the angles of a triangle is half a circle
Triangles can be broken up into right triangles so if we really understand
right triangles we can know them all.
The side opposite the right angle is called the hypotenuse
Certain ratios very important: sine, cosine, tangent
sin(A) = a/c
sin(A) = b/c
tan(A) = a/b = sin(A)/cos(A)
Additionally:
secant (sec) , cosecant (csc) , cotangent (cot) are reciprocals of these
(sec = 1/cos , csc = 1/sin, cot = 1/tan )
The fundamental theorem is the Pythagorean Theorem:
c a b
2
Proof:
2
2
Corollaries:
sin (  ) 2cos(  ) 21
for any angle 
Law of Sines
If a and c are two sides of a triangle opposite angles A and C then
a
c

sin(A) sin(C)
Proof: h = a sin(C) = c sin(A)
Law of cosines
If a, b, c are the sides of a triangle and C is the angle opposite c then
c2a 2b 22 a b cos(C)
c2( ba cos(C) ) 2a 2 sin(C) 2
b 22 a b cos(C)a 2 cos(C)2a 2 sin(C) 2
c2a 2b 22 a b cos(C)
Angle Measurement
Degree Measure: Divide circle into 360 parts.
Radian Measure:
Arc length is measure of the angle
Ratio of subtended arc length to radius = number of radiuses
"rad" = "radius"
radian measure 
length of arc
radius
Length of circumference =
2r
arc length  2  r 

 "radiuses"
radius
 r 
= 2  radians (rad)
360 degrees = 1 revolution (rev) = 2  rad
Unit conversions:
Write “1” in multiple ways
1=
360 deg
2  rad
1 rev
2  rad
=
=
=
2  rad
360 deg
2  rad
1 rev
Example: 5211 degrees = ____________ radians
(5211 deg) ( 1) = ( 5211 deg) (
2  rad
)=
360 deg
Assigning numbers to points on the unit circle
s -> P(s) = point on the circle “s” units counterclockwise , starting at
(1,0)
Identities
P( s2  )P( s )
sin( s2  )sin( s )
cos( s2  )cos( s )
From looking at right triangles we know that

sin  s cos( s )
2


cos s sin ( s )
2

From looking at the circle we see that
cos(s) = cos(-s)
sin ( s ) = sin( s ) = sin ( s )
Some identities aren’t so obvious
sin(A+B) = sin(A) cos(B) +sin(B) cos(A)
cos(A+B) = cos(A) cos(B) - sin(A) sin(B)
Where do these come from?
How calculators calculate sines, cosines, etc.
Some formulas you will learn in MA114.
1
1
1
1 5
1 6
ex1x x2 x3 x4
x 
x ...
2
6
24
120
720
1
1 5
1
1
sin ( x )x x3
x 
x7
x9...
6
120
5040
362880
1
1
1 6
1
cos( x )1 x2 x4
x 
x8...
2
24
720
40320
set xi  in ex
1
1
1
1 5
1 6
ex1x x2 x3 x4
x 
x ...
2
6
24
120
720
e
e
e
( i )
( i )
( i )
1
1
1
1 5 5
1 6 6
1i  i 2  2 i 3  3 i 4  4
i  
i  ...
2
6
24
120
720
1
1
1
1
1 6
1i   2 i  3  4
i  5
 ...
2
6
24
120
720
2 4
6
1
1 5
1  
...   3
 ...  i
2
24 720
6
120


e
( i )
cos(  )i sin (  )
set  = 
e
( i )
cos(  )i sin (  )
This produces, for instance:
e
( i )
-1
For now we learn Euler’s Identity as a memory device.
The following works when  1 ,  2
e
are real numbers
( i (    ) )
1
2
e
(i  )
1
e
(i  )
2
cos( 12 )i sin ( 12 )( cos( 1 )i sin ( 1 ) ) ( cos( 2 )i sin ( 2 ) )
cos( 12 )i sin ( 12 )
cos( 1 ) cos( 2 )sin ( 1 ) sin ( 2 )i ( sin ( 1 ) cos( 2 )sin ( 2 ) cos( 1 ) )
Comparing the real and complex parts we have
cos( 12 )cos( 1 ) cos( 2 )sin ( 1 ) sin ( 2 )
sin ( 12 )sin ( 1 ) cos( 2 )sin ( 2 ) cos( 1 )
Or
sin(A+B) = sin(A) cos(B) +sin(B) cos(A)
cos(A+B) = cos(A) cos(B) - sin(A) sin(B)
Setting A = B we get
sin( 2 A)2 sin( A) cos( A)
cos( 2 A )cos( A ) 2sin ( A ) 2
2
2
If we write sin ( A ) 1cos( A ) and substitute in the second and solve we
get
cos( A ) 2
1cos( 2 A )
2
and
sin ( A ) 2
1cos( 2 A )
2