Feb 11, 2011 The transformed trigonometric functions
Dec 30, 2015
What would be the amplitude:
• y = 2 cos x
• y = 8 sin 2x
• y = -3 cos x
• y = 2.4 sin 9x - 2
• amplitude = 2
• amplitude = 8
• amplitude = 3
• amplitude = 2.4
Another way to find amplitude:
Amplitude = half the distance
between the Max and min values= (M – m) 2= (2 - -6) 2
= 8 2= 4
Another way to find amplitude:
Amplitude = half the distance
between the Max and min values= (M – m) 2= (2 - -6) 2
= 8 2= 4
2
-6
What would be the value of a in the rule?
a = 1
Amplitude = half the distance
between the Max and min values= (M – m) 2= (2 - 0) 2
= 2 2= 1
In general then:
• For f(x) = a sin b(x – h) + k
OR:
• f(x) = a cos b(x – h) + k
Amplitude = |a|Max min
2
What would be the frequency:
• y = cos 4x
• y = 8 sin 2x
• y = -3 cos (x + 1) -2
• y = 2.4 sin (-9x) - 2
What would be the frequency:
• y = cos 4x
• y = 8 sin 2x
• y = -3 cos (x + 1) -2
• y = 2.4 sin (-9x) - 2
• frequency = 4
• frequency = 2
• frequency =
• frequency = 9
And if 4 cycles have a total width of 2.... ...then one of those cycles must have
a width of...
Amplitude
Period
Frequency
l.o.o.
y = sin 4x
And if 4 cycles have a total width of 2.... ...then one of those cycles must have
a width of...
Amplitude
Period
Frequency
l.o.o.
y = sin 4x
?
Amplitude
Period =
Frequency
l.o.o.
y = sin 4x
2
4
And if 4 cycles have a total width of 2.... ...then one of those cycles must have
a width of...
Amplitude
Period =
Frequency
l.o.o.
y = sin 4x
2
4
2
And if 4 cycles have a total width of 2.... ...then one of those cycles must have
a width of...
What would be the period:
• y = cos 4x
• y = 8 sin 2x
• y = -3 cos (x + 1) -2
• y = 2.4 sin (-9x) - 2
• period =
• period =
• period =
• period =
What would be the period:
• y = cos 4x
• y = 8 sin 2x
• y = -3 cos (x + 1) -2
• y = 2.4 sin (-9x) - 2
• period =
• period =
• period =
• period =
2
4 2
2
2
22
2
9
In general then:
• For f(x) = a sin b(x – h) + k
OR:
• f(x) = a cos b(x – h) + k
Frequency = |b|
Period = 2
b
But h does shift horizontally...and this shift has a special name:
Phase shift
Amplitude
Period
Frequency
l.o.o.
What would be the phase shift:
• y = cos 4x + 1
• y = 8 sin 2(x - ) -3
• y = -3 cos (x + 1) -2
• y = 2.4 sin (2x + )
• phase shift =
• phase shift =
• phase shift =
• phase shift =
What would be the phase shift:
• y = cos 4x + 1
• y = 8 sin 2(x - ) -3
• y = -3 cos (x + 1) -2
• y = 2.4 sin (2x + )
• phase shift = 0
• phase shift =
• phase shift = -1
• phase shift = 2
What would be the value of h in the rule?
If we consider this to be a sine function,
h = 2
Snake is beginning
here
What would be the value of h in the rule?
If we consider this to be a sine function,
h = 2
Which is /2 to the right of
where it usually begins
What would be the value of h in the rule?
If we consider this to be a sine function,
h =
In the rule, you would see:
2
x2
What would be the value of h in the rule?
If we consider this to be a cos function,
h =
Tulip is beginning
here
What would be the value of h in the rule?
If we consider this to be a cos function,
h =
Which is to the right of
where it usually begins
What would be the value of h in the rule?
If we consider this to be a cos function,
h =
Which is to the right of
where it usually begins
What would be the value of h in the rule?
If we consider this to be a cos function,
h =
In the rule, you would see:
(x - )
What would be the l.o.o.:
• y = cos 4x + 1
• y = 8 sin 2(x - ) - 3
• y = -3 cos (x + 1) - 2
• y = 2.4 sin (2x + )
What would be the l.o.o.:
• y = cos 4x + 1
• y = 8 sin 2(x - ) - 3
• y = -3 cos (x + 1) - 2
• y = 2.4 sin (2x + )
• l.o.o.: y = 1
• l.o.o.: y = -3
• l.o.o.: y = -2
• l.o.o.: y = 0
Another way to find k:
k = the number halfway between the Max and min
values= (M + m) 2= (1 + -3) 2
= -2 2= -1
Another way to find k:
k = the number halfway between the Max and min
values= (M + m) 2= (1 + -3) 2
= -2 2= -1
What would be the value of k in the rule?
k = the number halfway between the Max and min
values= (M + m) 2= (0 + -2) 2
= -2 2= -1
In general then:
• For f(x) = a sin b(x – h) + k
OR:
• f(x) = a cos b(x – h) + k
l.o.o. is the line y = k
Max mink
2
And another thing....
• For f(x) = a sin b(x – h) + k
OR:
• f(x) = a cos b(x – h) + k
Max = k + amplitudemin = k - amplitude
And another thing....
• For f(x) = a sin b(x – h) + k
OR:
• f(x) = a cos b(x – h) + k
Max = k + amplitudemin = k - amplitude
y = 2 sin 3(x - /4) + 1y
x
2
2
32
32
2
2
–2
–2
–
–
–32
–32
– 2
– 2
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
y = 2 cos 3(x + /4) + 1y
x
2
2
32
32
2
2
–2
–2
–
–
–32
–32
– 2
– 2
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2