Transcript 5.6 Graphs of Other Trig Functions

5.6
Graphs of Other Trig Functions
p. 602-603 1-12 all, 55-58 all
Review Table 5.6 on pg 601
Analysis of the Tangent Function
f  x   tan x
Domain: All reals except odd
multiples of  2
Range: , 

 3
2,3 2 by  4, 4
Continuous on its domain
Increasing on each interval in
its domain
Symmetry: Origin (odd function)
No Local Extrema
V.A.:
k
x
2
for all odd integers k

Unbounded
H.A.: None
End Behavior:
lim  tan x  and lim  tan x 
x 
x 
do not exist (DNE)
Analysis of the Tangent Function
f  x   tan x
sin x

cos x
 3
2,3 2 by  4, 4
What is the period
of the tangent function???
Period:

How do we know that these are
the vertical asymptotes?
They are where cos(x) = 0!!!
How do we know that these are
the zeros?
They are where sin(x) = 0!!!
Analysis of the Tangent Function
The constants a, b, h, and k influence the behavior of
y  a tan b  x  h  k
in much the same way that they do for the sinusoids…
• The constant a yields a vertical stretch or shrink.
• The constant b affects the period.
• The constant h causes a horizontal translation
• The constant k causes a vertical translation
Note: Unlike with sinusoids, here we do not use the
terms amplitude and phase shift…
Analysis of the Cotangent Function
cos x
f  x   cot x 
sin x
The graph of this function will
have asymptotes at the zeros
of the sine function and zeros
at the zeros of the cosine function.
2 , 2  by 4, 4
Vertical Asymptotes:
Zeros:
x
x
, 2 ,  ,0,  , 2 ,
3   3
,
, , ,
2
2 2 2
Guided Practice
Describe the graph of the given function in terms of a basic
trigonometric function. Locate the vertical asymptotes and
graph four periods of the function.
y   tan  2x 
Start with the basic tangent function, horizontally shrink by a
factor of 1/2, and reflect across the x-axis.
Since the basic tangent function has vertical asymptotes at all
odd multiples of  2 , the shrink factor causes these to move
to all odd multiples of  4 .
Normally, the period is  , but our new period is  2 . Thus,
we only need a window of horizontal length 2 to see four
periods of the graph…
Guided Practice
Describe the graph of the given function in terms of a basic
trigonometric function. Locate the vertical asymptotes and
graph four periods of the function.
y   tan  2x 
 ,  by 4, 4
Guided Practice
Describe the graph of the given function in terms of a basic
trigonometric function. Locate the vertical asymptotes and
graph two periods of the function.
f  x   3cot  x 2  1
Start with the basic cotangent function, horizontally stretch by
a factor of 2, vertically stretch by a factor of 3, and vertically
translate up 1 unit.
The horizontal stretch makes the period of the function 2 .
The vertical asymptotes are at even multiples of
.
Guided Practice
Describe the graph of the given function in terms of a basic
trigonometric function. Locate the vertical asymptotes and
graph two periods of the function.
f  x   3cot  x 2  1
How would you graph this with
your calculator?
y  3 tan  x 2 1
OR
2 , 2  by 10,10
y  3  tan  x 2    1
1
The graph of the secant function
1
y  sec x 
cos x
Wherever cos(x) = 1, its reciprocal sec(x) is also 1.
The graph has asymptotes at the zeros of the
cosine function.
The period of the secant function is
as the cosine function.
2 , the same
A local maximum of y = cos(x) corresponds to a
local minimum of y = sec(x), and vice versa.
The graph of the secant function
1
y  sec x 
cos x
1
2

1

2
The graph of the cosecant function
1
y  csc x 
sin x
Wherever sin(x) = 1, its reciprocal csc(x) is also 1.
The graph has asymptotes at the zeros of the sine
function.
 , the
The period of the cosecant function is 2
same as the sine function.
A local maximum of y = sin(x) corresponds to a
local minimum of y = csc(x), and vice versa.
The graph of the cosecant function
1
y  csc x 
sin x
1
2

1

2
Summary: Basic Trigonometric Functions
Function
Period
Domain
sin x
2
cos x
2
 , 
 , 
tan x
cot x


x   2  n
x  n
sec x
2
x   2  n
csc x
2
x  n
Range
1,1
1,1
 , 
 , 
 , 1 1, 
 , 1 1, 
Summary: Basic Trigonometric Functions
Function
Asymptotes
Zeros
Even/Odd
sin x
None
n
Odd
cos x
None
 2  n
Even
cot x
x   2  n n
x  n
 2  n
sec x
x   2  n
None
Even
csc x
x  n
None
Odd
tan x
Odd
Odd
Guided Practice
Solve for x in the given interval  No calculator!!!
sec x  2
3
 x
2
Let’s construct a reference triangle:
240
–1
 Third Quadrant
r
sec x  2 
x
r  2, x  1
Convert to radians:
60
2
4
x  240 
3
Guided Practice
Use a calculator to solve for x in the given interval.
3
csc x  1.5   x 
2
The reference triangle:
x
1
 Third Quadrant
r
csc x  1.5 
y
r  1.5, y  1
1
sin x  
1.5
1.5
x    sin
1
 2 3  3.871
Does this answer make sense with our graph?
Guided Practice
Use a calculator to solve for x in the given interval.
tan x  0.3 0  x  2
Possible reference triangles:
y
tan x  0.3 
x
y  0.3, x  1
-1
x
x
1
-0.3
0.3
x  tan
1
x    tan
 0.3  0.291
or
1
0.3  3.433
Whiteboard Problem
Solve for x in the given interval  No calculator!!!
3
sec x   2   x  2
5
x
4
Whiteboard Problem
Solve for x in the given interval  No calculator!!!
cot x  1   x  

2
3
x
4