Download TRIGONOMETRY AND CALCULUS Jan Murray

SUBAREA III – TRIGONOMETRY AND CALCULUS
Jan Murray-Abraham
CONNECTIONS BETWEEN RIGHT TRIANGLES, THE UNIT CIRCLE AND
TRIGONOMETRIC FUNCTIONS
The area of mathematics concerned with specific functions of angles and their application
to calculations is trigonometry, there are six functions of an angle commonly used in
trigonometry. They are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec),
and cosecant (csc) (Trigonometry). These functions are derived from the right triangle and
the unit circle.
Starting with a right triangle:
The six trigonometric functions are expressed as ratios of the 3 sides.
Cotangent, cosine and secant are the inverse of tangent, sine, and cosine respectively,
therefore:
Sin (_) =
opposite
b
=
hypotenuse c
Csc (_) =
Cos (_) =
adjacent
a
=
hypotenuse c
Sec (_) =
Tan (_) =
opposite b
=
adjacent a
Cot (_) =
1
sin e
=
hypotenuse c
=
opposite
b
hypotenuse c
1
=
=
adjacent
a
cos ine
1
adjacent a
=
=
tan gent
opposite b
The coordinates of any point on a circle can be found using the measure of an angle formed
by a right triangle. The hypotenuse of the right triangle is the radius of the circle. In a unit
circle the radius is one. Therefore:
sin _ =
y
=y
1
x
cos _ = = x
1
tan _ ==
y
x
From the unit circle above right cos _, sin _ and tan _ can be computed at any point on
this unit circle (Sparknotes):
cos 0 = 1
sin 0= 0
tan 0 = 0  (1,0)
cos !/2 = 0
sin !/2 =1
tan !/2 = undefined  (0,1)
cos ! =-1
sin ! = 0
tan ! = 0  (-1,0)
cos 3 !/2 = 0 sin 3 !/2 = -1
tan 3 !/2 = undefined  (0,-1)
The measure of angle _ that terminates in a quadrant is the shortest angular distance to the
x axis. This is called the reference angle. In the unit circle above right the angle of 30º in
the third quadrant is actually 180º +30º or 210º.
This means that 30º and 210º have the same value in terms of trigonometric functions
except that the sign (depending on the quadrant may be different).
3
For example tan 210 = +-tan 30 = +(using your calculator). Since 210 is in the third
3
3
quadrant, tan is positive (both x and y are both negative) so the answer is:
.
3
The theorem states:
A function of any angle will equal plus or minus that same function
of the corresponding acute angle.
The sign will depend on the quadrant.
To figure out the signs for the different functions in each quadrant remember that sin=y
and cos=x.
In the first quadrant _< 90º  x and y are both positive so ALL functions are positive.
In the 2nd quadrant 90º < _ > 180º  x is negative , y is positive so ONLY the sin and csc
are positive.
In the 3rd quadrant 180º < _ >270º  x and y are both negative so ONLY tan and cot are
positive.
In the 4th quadrant 270 º < _ > 360 º  x is positive y is negative so ONLY cos and sec are
positive.
Two Mnemonics that work well for keeping track of the signs and also the formulas for the
functions are:
1)All Students Take Calculus:
All are positive in 1st;
Sin is positive in 2nd;
Tan is positive in 3rd; and
Cos is positive in the 4th quadrant.
2) Some Old Horse Caught Another Horse Taking Oats Away:
sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent
USING DEGREE AND RADIAN MEASURES
Radians are another way to measure an angle. In the picture below the angle is equal to
the arc it sweeps on the circumference of the unit circle.
Since C=2!r, in a unit circle r=1, so C=2!. This means that the arc length swept by a
complete revolution of a unit circle is 360º. So:
360º =2! radians
1 radian = 360º /2! ≈ 57.3º
Since ! ≈ 3.14 then one complete revolution (2!) is about 6.28 radians.
Converting from degrees to radians:
Use 360º =2! radians or 180º/ ! = 1 radian
So 1 º = !/180º
Example:
a) convert 3 radians to degrees
b) convert 3º to radians
1 radian = 180º/ !
1º = !/180º
3 radians = 3*180º/ !
3º = 3 !/180º
=540º/ ! radians
= 0.05 radians
≈171.9º
The arc length spanned in a circle with radius r is: Arc length = r*_
ANALYZING CHARACTERISTICS OF THE GRAPH OF A TRIGONOMETRIC
FUNCTION (frequency, period, amplitude and phase shift)
Trigonometric functions are periodic functions- functions that repeat themselves at regular
intervals. So the graphs will look like waves. Radian measures are typically used in
graphing these functions. Here are some definitions for the characteristics:
Frequency is the number of cycles the function completes on an interval of 2! (360º)
Period is the time interval it takes a function to complete one full cycle.
Midline is half way between min and max.
Amplitude is the distance between the maximum or minimum value and the midline.
Phase shift is used to calculate the fraction of a full period that the curve has been shifted.
It tells the relative positions of two waves of the same period. (Connelly 2000)
Using the form:
y=AsinB(t - h) +k
|A|=amplitude
h = the horizontal shift
|B|= the frequency
|B|*h = phase shift
and
y=AcosB(t - h) +k
2!/ |B| = the period
y = k is the midline
The above graph shows how the change in frequency (B) changes the period (2! /|B|) but
the Amplitude is the same in both (Bourne).
The above graph shows the sin (x) with different amplitude (A) values and how that affects
the graph of sin x (Bourne). The frequency, thus the period does not change.
The graph below is an example of a horizontal shift. The dotted line is the ‘normal position
of the function y = sin 2x (Bourne). The blue line is the function y= sin(2x+1)
Putting y= sin(2x+1) in the form y = AsinB(t-h) +k
−1
y = sin 2(x-(
))
2
−1
so the horizontal shift = (
) ( negative means shifted left)
2
−1
phase shift = 2*(
) = -1
2
To find out how much the wave shifts, use this formula:
phase shift / 2! = -1/2! = _ of a full period shifted left
CONNECTIONS BETWEEN TRIGONOMETRIC FUNCTIONS AND POWER SERIES
One of the uses of a power series is to calculate trigonometric functions with out tables or
calculators. These were developed by Mathematicians including Newton in the late 17th
century.
One such series is the Taylor Series. The formulas for sine, cosine and tangent are as
follows:
e^x = a + bx + cx_+ dx_+……
= 1 + x + x2/2! + x3/3!+ x4/4! + ….
So:
sin x = x - x3/3! + x5/5! - x7/7! +...
cos x = 1 - x2/2! + x4/4! - x6/6! +...
tan x = x+ x3/3 + 2x5/15+ 17x7/315 + …
An important note is that the angle measure has to be in radians. Also, the numbers get
very close to zero very quickly so the first few calculations give you a good approximation.
Example:
What is sin 37º?
1º = !/180º  37º = 37!/180º  ≈ 0.645772
0.645772 = .645772
0.600889 = .645772 - .6457723/3!
0.60276 = .645772 - .6457723/3! + .6457725/5!
0.602742 = .645772 - .6457723/3! + .6457725/5! - .6457727/7!
0.602742 = .645772 - .6457723/3! + .6457725/5! - .6457727/7! + .6457729/9!
When calculated on the computer, sin 37º = .601815
This shows how rapidly the numbers converge toward zero. This series computes a good
estimation by only going through the first 5 calculations.
USING THE COMPLEX EXPONENTIAL FUNCTION TO EXPLORE PROPERTIES
OF TRIGONOMETRIC FUNCTIONS:
Since we know that:
ex = 1 + x + x2/2! + x3/3!+ x4/4! + ….
Then:
ex i =1 + ix + x2/2! +-i x3/3!+ x4/4! + i x5/5! + …
so:
isin x = ix - ix3/3! + ix5/5! - ix7/7! +...
cos x = 1 - x2/2! + x4/4! - x6/6! +...
ex = cos(x) + sin(x)
therefore Eulers formula:
ex i = cos(x) + isin(x)
Example:
If x = ! 
e iπ = -1 + 0 
e iπ + 1 = 0
From ex i = cos(x) = isin(x) comes the formulas:
(−1) n
e ix − e − ix
∑
2 n +1
2i
sin(x) = n =0 (2n + 1)! x
=
n
ix
−
∞ ( −1)
e + e ix
∑
2n
2
cos (x) = n =0 (2n)! x =
∞
ADDITIONAL RESOURCES
http://www.clarku.edu/~djoyce/trig/compute.html
http://www.explorelearning.com/index.cfm?method=cResource.dspCorrelatedStandards&
State=IL&Grade=11-12&Document=Performance%20Descriptors&Subj=Mathematics
http://en.wikipedia.org/wiki/Trigonometric_function
http://www.intmath.com/Trigonometric-graphs/1_Graphs-sine-cosine-amplitude.php
References
Bourne, M. ,Graphs of the Trigonometric Functions, retrieved April 18, 2007 from
http://www.intmath.com/Trigonometric-graphs/2_Graphs-sine-cosine-period.php
Connelly, Hughes - Hallett, Gleason. (2000). Functions Modeling Change. A Preparation
for Calculus. New York , NY: John Wiley & Sons, Inc.
SparkNotes, SAT Math Level 2, retrieved April 18, 2007
http://www.sparknotes.com/testprep/books/sat2/math2c/chapter9section1.rhtml
Spencer, L., The Math Page. (2001-2007), Trigonometry, retrieved April 17 from:
http://www.themathpage.com/aTrig/trigonometry-of-right-triangles.htm
Trigonometry. (2007). In Encyclopædia Britannica. Retrieved April 30, 2007, from
Encyclopedia Britannica Online: http://www.britannica.com/eb/article-9108709
Wicks, J. Exponential and Trigonometric Functions, retrieved April 20, 2007 from:
http://campus.northpark.edu/math/PreCalculus/Transcendental/Trigonometric/Defi
nition/index.html
PROBLEMS
1) You walk 4 miles around a circular lake. Give the angle in radians which represents
your final position relative to your starting point if the radius of the lake is 3 miles.
2) Using the picture find the height of the flag pole.
3) In a right triangle, csc_ = 13/5. Sketch the triangle and evaluate the remaining 5
functions.
4) State the amplitude, period, frequency, horizontal shift and phase shift of:
g(t) = cos(3t- π /4)
5) From the graph below what is the sin 330°?
6) The front door to the school gymnasium is 20 feet above ground level and is reached by a
flight of steps. To build a wheel-chair how long must the ramp be?
7) Find cos25° within 5 decimal places without using a calculator or table.
8) Using Euler’s formula if x = ! /2, compute ex i
SOLUTIONS
1) Using arc length (a) = r_ ; arc length is 4 and radius is 3
_= a/r so
_= 4/3≈1.33 radians
2) tan 37º= h/100
tan 37º* 100 = h
using your calculator .745*100=h
h=75.4 ft
3)
13
5
x
p
Using the Pythagorean Theorem: p_+5_=13_
sinx = 5/13
p_=169-15
cosx= 12/13
secx= 13/12
p_=144
tanx = 5/12
cotx = 12/5
p=12
4) Using the form AcosB(t-h)
g(t) = cos(3t- !/4)
= cos3(t- ! /12) so
amplitude = 1
frequency = 3
horitontal shift = !/12
5)
period = 2!/3
phase shift = !/4
π/4
= 1/8 of a period shift
2π
330° is the same as 30° in the 4th quadrant so sin30°= +- _. Since sin is
negative in the 4th quadrant then sin330°= -1/2
6)
20°
25
ground level
Using csc20°= h/25  h=csc20°*25
Sin20° = .342  csc20° = 1/.342
Length = 1/.342*25
= 73 ft.
7) Put into radians
25° = 25 π /180°
25° ≈ .43633
cos x = 1 - x2/2! + x4/4! - x6/6! +...
1 - .436332/2! = .90480
1 - . 436332/2! + . 436334/4! = .90632
1 - . 436332/2! + . 436334/4! - .436336/6! = .90631
cos25°≈ .90631
8)
ex i = cos(x) + isin(x)
= cos(! /2) + isin(! /2)
= 0 + i*1
=i