**8-2
Sine and Cosine Curves**
** **
*From
previous sections, we worked with horizontal and vertical stretching
and
shrinking. Recall that vertical stretching/shrinking happens by
changing
the amplitude of the graph. Horizontal stretching/shrinking
happens
by changing the period length of the graph. We will apply
these
concepts to the Sine and Cosine functions.*

* *
*For functions y = A
sin
Bx and y = A cos Bx (A not = 0 and B > 0)*
*amplitude
= | A | and period
= 2p/B*
* *
Demonstration
of Graph of y = sin Bx (Manipula Math)
__Graphing
Examples__
*1) Graph y = 3 sin 2x*

* *

*
Solution:
Remember that the basic sine curve has zeros at the beginning, middle
and
end of a cycle. Reaches its maximum at the 1/4 mark and has a
minimum
value at the 3/4 mark.*

*
This graph has an amplitude = 3 and a period length = 2p/2
= p*

*Thus, we have zeros at (0,
0), (p/2, 0) and (p,
0)*

*
maximum point at ( p/4, 3)*

*
minimum point at (3p/4, -3)*
*Plot these points
and
draw a smooth curve to get:*
*2) Graph y = (1/2)Cos
3x*

* *

*
Solution: Remember
that the basic cosine graph begins and ends at its maximum point.
In the middle, it is at its minimum value, and has zeros at the 1/4 and
3/4 mark.*

*
For this graph, the amplitude = 1/2*

*
period = 2p/ 3*

*
The maximums are at the beginning point (0, 1/2) and*

*
end point (2p/ 3, 1/2)*

*
minimum point at ( p/3, -1/2)*

*
Zeros at ( p/ 6, 0) and ( p/
2, 0)*
*Plot these points
and
smoothly connect to get:*
*3) Graph y = -2 Sin (p/2)x*

* *

*
Solution: The
amplitude = | -2 | = 2*

*
Period = 2p/ (p/
2) = 4*

*
Note that this graph is a reflection about the x-axis. This
interchanges
the maximum and minimum values.*

*
zeros at (0, 0), ( 2, 0), ( 4, 0)*

*
minimum ( 1, -2)*

*
maximum ( 3, 2)*
*Plot these points
and:*
*4) Graph y = -3 cos (p/
5)x*

* *

*
Solution: This
again is a reflection about the x-axis.*

*
amplitude = | -3 | = 3*

*
period = 2p/ (p/
5) = 10*

*
Minimum values at (0, -3), (10, -3)*

*
Maximum value at ( 5, 3)*

*
Zeros at ( 2.5, 0) and ( 7.5, 0)*
*Connect the points
and:*

__Determining
the equation from the graph__
** **

**1) ***Find the amplitude and
period length of each function. Then write an equation.*
*
Solution: The graph is a sine curve
because it begins at (0, 0). The amplitude = 3 and the period
length
is (1/2)p.*

*
Since the amplitude = 3, A = 3.*

*
Because period = (1/2)p, B = 2p/(1/2)p
= 4*

*
Thus, the equation is: y = 3 sin 4x*

* *

*2) Find the amplitude
and period for the function. Write an equation for the graph.*

*Solution:
amplitude = 2 and period length = 1*

*
The graph is a cosine graph reflected about the x-axis. The graph
starts at a minimum.*

*
A = -2 because it is reflected about the x-axis.*

*
B = 2p/1 = 2p*

*
Thus the equation is: y = -2 cos 2px*

__Solving
an equation algebraically or graphically__
* *

*1) Solve the equation
for 0 *__<__ x < 2p. 3
sin
2x = 1

*
a) Graphically using the TI-82 calculator*

*
Set calculator to radian mode. Go to y= screen.*

*
Type 3 sin 2x for y*_{1}
and 1 for y_{2}
Press graph. Zoom in.
*Notice that the
graphs
cross four times between 0 and 2p.
Using
your calculator, press the calc button, then the intersect
button.
Guess first curve, second curve, then best guess. Answer to the
point
furthest to the left is :*
*(.17, 1)
Repeat
the process for the other 3 points.*
*Other answers
are:
(1.40, 1), (3.31, 1) and (4.54, 1).*
* *
*
b) Algebraically.*
* *
*3 sin 2x = 1*
*sin 2x = 1/3*
*2x = sin*^{-1
}(1/3)
*The reference angle
is
.34 (using your calculator)*
*We need to check
through
2 periods to find all four answers.*
*sine is also
positive
in quad II (3.14 - .34) = 2.80*
*Now add 6.28 to
each
of these values to get all four answers.*
*2x = .34, 2,80,
6.62,
9.08*
*x = .17, 1.40,
3.31,
4.54*

**In
the next section, we will add translations to the graph of sine and
cosine!!**