Download Chapter 5: Trigonometric Functions of Real Numbers

Lecture Notes
Algebra II Honors
Chapter 5: Trigonometric Functions of Real Numbers
5.1 The Unit Circle
We will define the trigonometric functions as functions of real numbers, represented as arc
lengths along a unit circle. The unit circle has center at the origin and radius one. Its equation is:
x2 + y2 = 1.
Example: A point on the unit circle in quadrant I has the same x and y coordinates. Find them.
Since the point P(x, y) id on the unit circle, its coordinates must satisfy x2 + y2 = 1. But we are
given that y = x. Thus 2x2 = 1or x = 1 / √2 and so y = 1 / √2. This is marked above as
⎛ 2 2⎞
,
⎜⎜
⎟
2
2 ⎟⎠
⎝
Terminal Points: At the point (1, 0) on the unit circle above, imagine a vertical real number line
tangent to the circle with 0 on the number line at the point of tangency. Wrap this line about the
circle. Positive numbers wrap CCW, and negative numbers wrap CW. Identify the terminal point
P(x, y) on the unit circle as the point corresponding to real number line value t.
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Algebra II Honors
An applet that illustrates this is at can be found here:
http://members.shaw.ca/ron.blond/TLE/WRAPPING.FUNCTION.APPLET/index.html
The circumference of a circle is 2πr, where the radius r = 1 for the unit circle. The unit circle thus
has circumference 2π, and the following is the correspondence between points on the circle and
the number line being wrapped. Remember to start with t = 0 at P(1, 0).
(1, 0) corresponds to t = 0 or 2π or 4π or ... and also –2π or –4π or ...
(0, 1) is 1/4 the way around the circle so corresponds to 1/4 of 2π or t = π/2.
Other points are marked in the diagram above. We show below how to find the terminal point
corresponding to t = π/6.
From the point P(x, y) on the unit circle in the first quadrant that corresponds to t = π/6, draw
segments to R(0,1) and Q(x, –y). Note both of these arcs have length π/3, and so the segments
(chords) are equal. Thus the lengths PR = PQ or
2y =
( x − 0 ) + ( y − 1)
2
2
→ 4y 2 = x 2 + y 2 − 2y + 1
Since on the unit circle x2 + y2 = 1, we have 4y2 + 2y – 2 = 0 or 2(2y2 + y – 1) = 0. The point can
−1 ± 12 − 4 ⋅ 2 ⋅ −1 −1 ± 3 1
=
= or − 1 , but only y = 1/2 is in the first quadrant.
2⋅2
4
2
⎛ 3 1⎞
Substituting this for y in x2 + y2 = 1, we find P ⎜
⎜ 2 , 2 ⎟⎟ .
⎝
⎠
have y =
How do we find the coordinates of terminal points that are not in quadrant I?
The reference number t associated with t is the shortest distance along the unit circle between the
terminal point P(x, y) determined by t and the x-axis. Thus 0 ≤ t ≤ 2π, and we can find a terminal
point on the unit circle in quadrant I (call it Q(a, b)) determined by t . Then the coordinates of P
are the same as the coordinates of Q to within ±. Choose the correct sign for the quadrant of P.
Example: Find the terminal point determined by t = 5π/6.
P(x, y) is in II since π/2 < 5 π/6 < π. Along the unit circle from P(x, y) to (–1, 0) we find
t = π – 5 π/6 = π/6 is the reference number. We have shown above that the terminal point Q(a, b)
⎛ 3 1⎞
in I determined by t is Q ⎜
⎜ 2 , 2 ⎟⎟ . Since P(x, y) in II has the same coordinates to within ±, by
⎝
⎠
⎛
3 1⎞
symmetry we have P ⎜ −
⎜ 2 , 2 ⎟⎟ .
⎝
⎠
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Algebra II Honors
5.2 Trigonometric Functions of Real Numbers (Circular Functions)
Let t be any real number and let P(x, y) be the terminal point on the unit circle determined by t.
•Define six trig functions as below. Note we cannot divide by zero, so some values are undefined.
sin(t) = y
cos(t) = x
tan(t) =
y
x
1
y
1
sec(t) =
x
x
cot(t) =
y
csc(t) =
•Example: Find all six trig functions of t = 5π/6
⎛
⎜
⎝
As we saw in the last section, t = 5π/6 determined the terminal point P ⎜ −
3 1⎞
, ⎟.
2 2 ⎟⎠
Using the above definitions we find:
⎛ 5π ⎞ 1
sin ⎜ ⎟ =
⎝ 6 ⎠ 2
⎛ 5π ⎞
csc ⎜ ⎟ = 2
⎝ 6 ⎠
3
⎛ 5π ⎞
cos ⎜ ⎟ = −
2
⎝ 6 ⎠
2 3
⎛ 5π ⎞
sec ⎜ ⎟ = −
3
⎝ 6 ⎠
3
⎛ 5π ⎞
tan ⎜ ⎟ = −
3
⎝ 6 ⎠
⎛ 5π ⎞
cot ⎜ ⎟ = − 3
⎝ 6 ⎠
•Note the domains of the sin and cos are all real numbers, but the others are restricted.
tan and sec have as domains all reals except for π/2 + n·π for all integer n
cot and csc have as domains all reals except for integral multiples of π.
•Note All trig functions are positive in quadrant I, only the sin and csc in II, the tan and cot in III,
and the cos and sec in IV. Memorize as All Students Take Calculus.
•Note by virtue of the definitions, csc and sin are reciprocals. Also sec and cos, and tan and cot.
•Note the cos and sec are even functions, while the others are all odd.
•Identities
Reciprocal:
csc(t) =
1
sin(t)
sec(t) =
1
cos(t)
cot(t) =
1
tan(t)
Quotient:
tan(t) =
sin(t) sec(t)
=
cos(t) csc(t)
Pythagorean:
sin 2 ( t ) + cos 2 ( t ) = 1
cot(t) =
cos(t) csc(t)
=
sin(t) sec(t)
1 + tan 2 ( t ) = sec 2 ( t )
1 + cot 2 ( t ) = csc2 ( t )
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Lecture Notes
Algebra II Honors
•Example. Given sin(t) = 3/5, t in II. Find the other trig functions.
Recall that sin(t) is defined as the y coordinate of a point P on the unit circle. Thus y = 3/5 and
since x2 + y2 =1, x = ±4/5. In II, x (and hence cos(t)) are negative and so cos(t) = –4/5.
Now csc(t) = 1/sin(t) = 5/4 and sec(t) = 1/cos(t) = –5/3.
Using the definition tan(t) = y/x, we find tan(t) = –3/4 and cot(t) = –4/3.
The student may ask how these definitions square with those learned in geometry for the trig
functions. From the terminal point P drop a perpendicular to the x-axis and construct the right
triangle shown below. Recall the geometry definitions in terms of the ratio of sides of a right
triangle, and note the hypotenuse is one (unit circle).
For example, sin ( θ ) =
opposite
y
= = y . All the other SOHCAHTOA definitions follow.
hypotenuse 1
The angle θ has the same dimensionless magnitude as the arc length t from (1, 0) to P(x, y) if we
measure θ in radians. One radian is the central angle of a circle that subtends an arc equal to its
radius. One revolution is equivalent to an arc equal in length to the circumference. Thus 1
revolution equals 2π radians equals 360°.
θ=
t
where t is arc length, r is radius, and θ is measured in radians.
r
For the unit circle, r = 1 and θ = t.
Example: Find sin(1.3). Set calculator to radian mode and find sin(1.3rad) = 0.9636
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5.3 Trigonometric Graphs
Recall that sin(t) is the y coordinate of the terminal point P(x, y) on the unit circle determined by
the real number t. We found that if the central angle θ in the diagram above is measured in
radians, it has the same magnitude as the arc it subtends (t).
Thus as t increases, θ increases, and P moves CCW about the unit circle. Values of the
y-coordinate of P are plotted for various t or θ. Since the circumference of the unit circle is 2π,
these values repeat with periodicity 2π. Since the unit circle has radius one, sine and cosine
graphs have |y| ≤ 1. The basic sine graph is shown above.
In the same way we can construct the graph of cos(t) using the x-coordinates of P.
1.0
0.5
1
2
3
4
5
6
-0.5
-1.0
Both sine and cosine graphs have period 2π. Thus:
sin(t + 2n π) = sin(t), for any integer n.
cos(t + 2n π) = cos(t), for any integer n.
Sinusoidal graphs are graphs of the sine and cosine subject perhaps to transformations such as
vertical shifts, horizontal shifts, vertical stretching or shrinking, and horizontal stretching or
shrinking. Graphs showing such transformations are in the Mathematica worksheet "Trig Graphs"
at http://algebra2.home.comcast.net/~algebra2/Mathematica/Trig%20Graphs.pdf.
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Algebra II Honors
Our goal is to quickly sketch graphs of the form y = A · sin[B · (x – C)] + D, for nonzero
constants A, B, C, and D. All written below applies equally well to the cosine graph.
•Amplitude (A): y = A · sin(x) is a vertical stretch (|A| > 1) or shrink (0 < |A| < 1) of the basic sine
graph. The amplitude is always taken as |A| in such equations. If A is negative, the graph is a
reflection in the x-axis of the usual sine graph. Below are graphs of y = 2sin(x) in red and (in
blue) y = sin(x). Note the 2sin(x) graph goes 2 above and below the horizontal axis instead of 1.
Amplitude
2
1
1
2
3
4
5
6
-1
-2
•Period or Frequency (B): y = sin(B · x) is a horizontal stretch (0 < B < 1) or shrink (B > 1) of the
basic sine graph. The usual period of sine is 2π; these graphs have period 2π / B. Thus instead of
one complete wave per 2π, these graphs have B complete waves. The frequency or number of
complete waves per 2π interval is B. Note if B is negative one may use the odd/even properties of
sine/cosine to factor this out, leading to possible x-axis reflections as shown above.
Below are graphs of y = sin(2x) in red and (in blue) y = sin(x). Note the sin(2x) graph has 2
complete waves or cycles or oscillations in 2π - the period of sin(2x) is 2π / 2 = π.
Period
1
0.5
1
2
3
4
5
6
-0.5
-1
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Algebra II Honors
•Phase Shift (C): y = sin(x – C) is a horizontal shift of C units (right if C > 0, left if C < 0) of the
basic sine graph. Below are graphs of y = sin(x – π/2) in red and y = sin(x) in blue. Note the red
graph is a typical sine graph that has been shifted π/2 or 1.57 units to the right.
Phase
Shift
1
0.5
1
2
3
4
5
6
-0.5
-1
•Vertical Shift (D): y = sin(x) + D is a vertical shift of D units (up if D > 0, down if D < 0) of the
basic sine graph. Below are graph s of y = sin(x) + 1 in red and y = sin(x) in blue. Note the red
graph is shifted upward by 1 unit.
Vertical
Shift
3
4
2
1.5
1
0.5
1
2
5
6
-0.5
-1
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Lecture Notes
Algebra II Honors
•These transformations can be combined in one graph of y = 2 · sin[2 · (x – π/2)] + 1
Be sure the B constant is factored outside of (x – C): that is, 2x – π → 2(x – π/2)
y =2*sin@2*Hx - pê2LD + 1
3
2
1
1
2
3
4
5
6
-1
One final note: when graphing trig functions on the TI-86, a good viewing window or rectangle
may be obtained by choosing Zoom → ZTrig.
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Algebra II Honors
5.4 More Trigonometric Graphs
•Graphing y = sec(x) and y = csc(x)
Since these functions are reciprocals of the cosine and sine graph, respectively, for each x plot the
reciprocal of the cosine or sine y coordinates. This reveals vertical asymptotes in the graphs of
sec and csc wherever cos or sin have y = 0. The vertical asymptotes are shown on the graphs
below but are not part of the sec or csc graph. On the TI-86, select Dot mode to draw the graphs.
Sec@xD vs x
5
1
2
3
4
5
6
4
5
6
-5
Csc@xD vs x
8
6
4
2
1
2
3
-2
-4
-6
Sec(x) and csc(x) have period 2π like cos(x) and sin(x) and transformations are handled in the
same way. Note sec and csc avoid the zone between y = –1 and 1.
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•Graphing tan(x) and cot(x)
These functions have period π, as opposed to 2 π for the other functions.
Thus the period of tan(B · x) or cot(B · x) is π/B, as opposed to 2π/B.
Other transformations are handled in the usual manner.
The amplitude of these functions is not much help in graphing: more useful is noting that both
have y = 1 for x = π/4.
Tan@xD vs x
6
4
2
1
2
3
4
5
6
4
5
6
-2
-4
-6
Cot @xD vs x
6
4
2
1
2
3
-2
-4
-6
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Algebra II Honors
5.5: Modeling Harmonic Motion
The great utility of sine and cosine functions is their periodicity, enabling them to serve as
mathematical models for physical systems that are periodic. We examine below the periodic
vibrations of a mass on a spring.
A mass m hangs from a vertical spring of spring constant k. Recall k is the force (in newtons)
needed to stretch the spring one meter. A stiff spring has a large k, a slinky has very small k.
We observe the oscillating mass. At time t = 0, it is at y = 0 and heading upward. It reaches a
maximum positive y value of y = A (recall |A| is the amplitude of the motion). If there is no air
resistance or damping, the mass will continue to oscillate between ±A forever.
⎛ k ⎞
⎜ m ⋅ t ⎟⎟ models the periodic motion.
⎝
⎠
The equation y = A ⋅ sin ( ω⋅ t ) = A ⋅ sin ⎜
The choice of the sine function is related to the starting position of the mass at t = 0. If the mass
was at y = ±A at t = 0, a cosine function could be used. Any starting position at t = 0 may be
modeled with a sine or cosine function with phase shift and/or vertical shift.
Those functions would be of the form: y = A ⋅ sin ⎡⎣ω ( t − C ) ⎤⎦ + D
The period of the oscillation is 2π / ω and there will be ω complete oscillations every 2π seconds.
The frequency f is the number of oscillation per second; it is ω / 2π = 1 / period.
An example is worked on the next page.
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Algebra II Honors
Example: The displacement of a mass suspended from a spring is modeled by y = 10 · sin (4π · t),
where y is measured in inches and t in second. Find the amplitude, period, frequency, and graph.
Compare y = 10 · sin (4π · t) to the equation y = A · sin(ω · t).
Amplitude = |A| = 10 inches
Period = 2π / ω = 2π / 4 π = 1/2 second.
Frequency = 1 / period = 2 Hz or cycles per second
y = 10*Sin@4*p*t D
10
5
0.2
0.4
0.6
0.8
1.0
-5
-10
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