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Trigonometry
Notes on the Tangent and Cotangent Functions and Their Graphs.
The Basic Tangent Function: First because tan(x) =
sin(x)
cos( x)
, tan(x) will be undefined when cos(x) = 0. Moreover, since sin(x) = 1 or -1 when cos(x) = 0, y = tan(x)
has vertical asymptotes when cos(x) = 0.
Plus, tan ( x + π ) = tan ( π2 − ( -x − π2 ) ) = cot ( -x − π2 ) = cot ( - ( x + π2 ) ) = - cot ( x + π2 ) = - cot ( π2 − ( -x ) ) = - tan ( -x ) = tan ( x ) . Thus,
y = tan(x) has a period of π. Since we would like to have the basic period to be continuous, we set the basic period to be ( - π2 , π2 ) to
be between 2 consecutive vertical asymptotes.
Therefore, we get:
The Basic Tangent Function
y = tan(x)
3π
5π
7π ⎫
π
⎧
Domain = ⎨ x | x ≠ ± , ± , ± , ± ...⎬
2
2
2
2 ⎭
⎩
Range = ( -∞, ∞ )
Period = π
⎛ π π⎞
Basic Period = ⎜ - , ⎟
⎝ 2 2⎠
Vertical Asymptotes at
π
3π
5π
7π
x = ± , ± , ± , ± ...
2
2
2
2
Odd function: Symmetric to origin
tan(-x) = -tan(x)
Graphing Criterion:
At least 1 period,
2 vertical asymptotes as dashed lines,
3 plotted points including:
inflection point(where it changes how it’s bending).
The General Tangent Function
y = a tan(bx+c)+d where b > 0
{ x |cos(bx + c) ≠ 0 }
( -∞,∞ )
Domain =
Range =
Period =
π
b
Phase Shift = -
c
b
⎡ -π − 2c π − 2c ⎤
,
Basic Period moved to = ⎢
2b ⎥⎦
⎣ 2b
Vertical Shift = d
Graphing Criterion:
At least 1 period,
2 vertical asymptotes as dashed lines,
3 plotted points including:
inflection point(where it changes how it’s bending).
SCC:Rickman
Notes on the Tangent and Cotangent Functions and Their Graphs.
Page #1 of 3
The Basic Cotangent Function: Through a similar method we can get the graph of y = cot(x) =
cos(x )
sin( x )
.
The Basic Cotangent Function
y = cot(x)
Domain = {x | x ≠ 0, ± π, ± 2π, ± 3π, ± 4π ...}
Range = ( -∞, ∞ )
Period = π
Basic Period = ( 0,π )
Vertical Asymptotes at
x = 0, ± π, ± 2π, ± 3π, ± 4π ..
Even function: Symmetric to y-axis
cot(-x) = cot(x)
Graphing Criterion:
At least 1 period,
2 vertical asymptotes as dashed lines,
3 plotted points including:
inflection point(where it changes how it’s bending).
Graphing Transformations of Tangent and Cotangent: To keep from having to remember all the different formulas for each of the different trigonometric functions, we’ll just focus on
using the inequality to find where the basic period moves.
π⎞
⎛
Example #1: Graph y = 2 tan ⎜ x + ⎟ + 4 . Identify period, phase shift and vertical shift.
3⎠
⎝
►
First, find where the basic period has moved to:
π
π π
- <x+ <
2
3 2
3π 2π
3π 2π
- −
<x<
−
6
6
6
6
5π
π
<x<
6
6
5π π
π ⎛ 5π ⎞ 6 π
= π , and the basic period will have asymptotes at x = - , .
Thus, the period is − ⎜ - ⎟ =
6 ⎝ 6 ⎠ 6
6 6
5π
π
- +
π
-5π+π - 4π
=
=- .
The inflection point will be halfway between the asymptotes. Thus, at x = 6 6 =
2
12
12
3
While the points halfway between the inflection point and each asymptote will be stretched by a factor of 2. Since they start at
( - π4 , -1) and ( π4 ,1) , they will be 2 below and 2 above respectively from the baseline.
Vertical Reflection: No
π
Period = = π
1
π
PhaseShift = 3
⎡ 5π π ⎤
Basic Period moved to ⎢ - , ⎥
⎣ 6 6⎦
VerticalShift = 4
□
SCC:Rickman
Notes on the Tangent and Cotangent Functions and Their Graphs.
Page #2 of 3
Notice that the only formula to change from the previous 4 trigonometry functions is the one for the period. Assuming b is positive,
π
it’s now Period = for y = a tan(bx+c)+d or y = a cot(bx+c)+d.
b
Example #2: Graph y = 3cot ( π − 2x ) . Identify period, phase shift and vertical shift.
►
y = 3cot ⎡⎣ - ( 2x − π ) ⎤⎦
y = -3cot ( 2x − π )
0 < 2x − π < π
π < 2x < 2π
π
<x<π
2
Vertical Reflection: Yes
π
Period =
2
π
PhaseShift =
2
⎡π ⎤
Basic Period moved to ⎢ , π ⎥
⎣2 ⎦
VerticalShift = 0
□
Applications: Example #3: If a plane is flying at a constant altitude of 8000m directly over an observer, assuming the Earth is flat a) find the
relation between the horizontal distance from the observer to the plane and the angle of elevation from the observer to the plane and ,
b) graph it.
►
a)
tan ( θ ) =
opp 8000
=
adj
x
x = 8000 cot ( θ )
b) Since angle of elevation would be an acute angle and the plane would never be on the ground for this problem, the domain would
⎛ π⎤
be ⎜ 0, ⎥ . Therefore the graph would be
⎝ 2⎦
□
SCC:Rickman
Notes on the Tangent and Cotangent Functions and Their Graphs.
Page #3 of 3