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Trigonometric functions wikipedia , lookup

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```Graphs of Trigonometric Functions
Here are the graphs of the six main trigonometric functions along with
their domains and ranges. The windows shown have x-values from
−2𝜋 to 2𝜋, and y-values from -3 to 3:
y  sin x
Domain:
Range:
Period:
All Real Numbers
[-1, 1]
2𝜋
y  cos x
Domain:
Range:
Period:
All Real Numbers
[-1, 1]
2𝜋
y  tan x
Domain:
All Real Numbers
𝜋
except 2 ± 𝑘𝜋
Range:
Period:
y  csc x
Domain:
Range:
(-∞, ∞)
𝜋
All Real Numbers
except :0 ± 𝑘𝜋
(-∞, -1]
and [1, ∞)
Period:
y  sec x
Domain:
Range:
Period:
y  cot x
Domain:
Range:
Period:
2𝜋
All Real Numbers
𝜋
except 2 ± 𝑘𝜋
(-∞, -1]and[1,∞)
2𝜋
All Real Numbers
Except 0 ± 𝑘𝜋
(-∞, ∞)
𝜋
TransformationsThe general form for the sine equation is:
(In our parent graph: y  sin x ;




y  a sin(bx  c )  d
a = b = 1, c = d = 0)
Changing the value of ‘a’ changes the amplitude of the graph.
‘a’ is the amplitude of the graph.
Changing the value of ‘b’ changes the period of the graph. The
new period is found by taking the original period and dividing by
‘b’
Changing the values of ‘b’ and ‘c’ creates phase shift. This
c
moves the graph horizontally. The phase shift is equal to: 
b
Changing the value of ‘d’ creates vertical displacement. If ‘d’ is
positive, the graph will be moved up. If ‘d’ is negative, the
graph will be moved down.
Example Set 1:
Find the amplitude, period, phase shift and vertical displacement of
these graphs:
a)
y  3sin(4 x  20)  5
b)
y
1
1
cos( x  4)  1
2
2
c)
𝜋
𝑦 = tan⁡(2𝑥 − 2 )
Example Set 2:
Give the equation for the sine function that has the following
characteristics:
a)
Amplitude =
Period =
2
b)
Amplitude =
Period =
8𝜋
2𝜋
5
Phase Shift =
5𝜋
½
𝜋
Phase Shift = − 4
9
Example Set 1:
a
3
Amplitude
Period
Phase Shift
5
Vertical
Displacement
+5
Note:
b
½
𝜋
2
c
n/a (See Note)
𝜋
2
𝜋
4
None
4𝜋
-8
-1
We don’t talk about amplitude for tangent, as this graph
goes off to infinity in both directions.
Period Calculations:
a)
2𝜋
4
=
𝜋
2
b)
2𝜋
1
2
= 4𝜋
c)
𝜋
2
=
𝜋
2
Phase Shift Calculations:
a)
c
20
 
5
b
4
b)
c
4
    8
1
b
2
c)
𝑐
−𝑏 = −
−
𝜋
2
2
=
𝜋
4
Example Set 2:
a)
𝑦 = 2sin⁡(5𝑥 −
25𝜋
9
)
1
b)
𝑥
𝜋
𝑦 = 2 sin⁡(4 + 16)
Calculations to find ‘b’:
Recall that the new period is the old period divided by b:
a)
2𝜋
5
=
𝑏=
2𝜋
𝑏
2𝜋
b) 8𝜋 =
2𝜋
5
2𝜋
𝑏
2𝜋
𝑏 = 8𝜋
1
𝑏=5
𝑏=4
Calculations to find ‘c’:
We find ‘c’ by using the phase shift as well as the value we just found
for ‘b’.
c
Remember that the phase shift is equal to: 
b
𝑐
a)
𝑃𝑆 = − 𝑏
5𝜋
9
𝑐
= −5
𝑐=
−25𝜋
9
b)
𝑐
𝑃𝑆 = − 𝑏
𝜋
𝑐
−4 = − 1
4
𝜋
𝑐 = 16
```