θ
θ
π
θ
The function y = 3 sin
has the same cycle length as y = sin
, namely 2
. but y = 3 sin
rises to
θ
π
π
maximum value of 3 (when
=
2
, for instance) and falls to a minimum of -3 (at
3
2
, for
instance). Its graph is shown below in Figure T27.
Figure T27
The maximum height above the horizontal axis of a graph that oscillates equally above and below the
x-axis is called the amplitude. Thus, y = sin x has amplitude 1, y = 2 cos x has amplitude 2, and y = -2
sin x has amplitude 2.
θ
θ
π
θ
Now consider the function y = sin 2
. Its amplitude is 1, but because 2
runs from 0 to 2
while
is
π
π
running from 0 to
, this function completes one cycle in
units. At a glance its graph looks exactly
like the sine function. The only difference is that it cycles faster (twice as fast) as the ordinary sine
function. Its graph is shown below in Figure T28.
Figure T28
π
The period of this function is
, which is the length of one cycle. Now we combine these ideas. A
θ
θ
function of the form y = A sin B
or y = A cos B
has
π
2
Amplitude = |A| and Period =
.
B
Problems:
θ
40. Find the amplitude and period of the function y = 3 sin 2
.
θ
41. Find the amplitude and period of the function y = -2 cos 5
.
θ
=
42. Find the amplitude and period of the function
y
4
cos
.
3
43. Find the amplitude and period of the function shown in Figure T29. Find the equation for the
function.
π
6
Figure T29
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