1. Aim: What’s the a in y = a sin x all about?
Do Now:
Graph the following:
set your calculator window to following settings:
Xmin= -1
Xmax=2.5
Xscl=/2
Ymin=-4
Ymax=4
Yscal=1
then graph the following
y = sinx; y = 2 sinx; y = 3 sinx

2. y = a sin x y = sin x amplitude - 1 y = 2 sin x amplitude - 2-1
maximum
minimum π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
y = sin x
amplitude - 1
maximum
minimum
y = 2 sin x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 2 -2
amplitude - 2
y = 3 sin x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 3 -3
maximum
minimum
amplitude - 3

3. y = a cos x y = cos x amplitude - 1 y = 2 cos x amplitude - 2
radians
y = cos x 1 -1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
amplitude - 1
y = 2 cos x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 2 -2
amplitude - 2
y = 3 cos x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 3 -3
amplitude - 3

4. In general, for the functions
What the a is All About
y = a sin x
y = a cos x
The amplitude of a periodic function is half
the difference between the minimum and
maximum values of the function.
In general, for the functions
y = a sin x and y = a cos x:
amplitude = | a |
max = 6
min = -6
ex.
y = 6 sin x
6 – (-6) 2
= 6
amplitude is 6

5. Sketch the graph of y = -2 cos x over the interval 0 ≤ x ≤ 2π:
What if a is negative?
Sketch the graph of y = -2 cos x over the interval 0 ≤ x ≤ 2π:
(w/o graphing calculator)
1. Table of values
2. Plot points & sketch x
-2cos x = y
-2cos 0 = -2
π/2
-2cos π/2 = π
-2cos π = 2
3π/2
-2cos 3π/2 = 2π
-2cos 2π =
Why?
y = 2 cos x 2
y = -2 cos x
amplitude = 2 1
π/2 π
3π/2 2π
5π/2 -1 -2
key points Min.,
Zero, and Max.

6. y = a cos x & y = (-a)cos x are reflections
What if a is negative?
y = 2 cos x 2 1
π/2 π
3π/2 2π
5π/2 -1 -2
y = -2 cos x ?
y = 2 cos x
y = -2 cos x
rx-axis
y = 2 cos x
y = -2 cos x
y = a cos x & y = (-a)cos x are reflections
of each other through the x-axis.
y = a sin x & y = (-a)sin x are reflections
of each other through the x-axis.

7. What about the b? y = cos x y = cos 2x y = cos 1/2x y = cos 4x 1 -1 1π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 1 -1
y = cos x -1 1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
y = cos 2x -1 1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
y = cos 1/2x -1 1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
y = cos 4x

8. What about the b? y = sin x y = sin 2x y = sin 1/2x y sin 4x 1 -1 1 -1π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 -1 1
y = sin 2x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 -1 1
y = sin 1/2x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 -1 1
y sin 4x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2

9. 1 period – 2π 2 1 2 What about the b? y = sin x y = sin 2x-1
y = sin x 1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
period – 2π -1 1
y = sin 2x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 2
period – π
How often does the cycle repeat itself over
the interval 0 ≤ x ≤ 2π? 1
y = sin x
one time 2
y = sin 2x
two times
frequency (b) – of a periodic function is the
number of cycles from 0 ≤ x ≤ 2π. (the number
of times the function repeats itself).

10. 1 1 cycle 2 2 cycles Length of cycle? y = sin x y = sin 2x In General:-1
y = sin x 1 π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2
1 cycle -1 1
y = sin 2x π 2π
3π/2
π/2 3π 4π
7π/2
5π/2 5π
9π/2 2
2 cycles
In General:
number of cycles from 0 to 2π
• length of 1 cycle
= 2π b
• period
= 2π
of a function
= 2π
|b|
period
= (2π)
1/3
pd.
= 6π
ex. y = sin 1/3x
b = 1/3

11. Understanding Sine/Cosine Curves
y = a sin bx
y = a sin bx
of a function
= 2π
|b|
period
amplitude = | a |
frequency = |b|
Sketch the graph of y = 3 sin 2x in the
interval 0 ≤ x ≤ 2π.
1) Determine the amplitude & period
a = 3, b = 2
amplitude = | a | = 3
= 2π
|b|
period
|2|
= π
max. = 3
min. = -3
divide the period π,
into 4 equal intervals: π/4, π/2, 3π/4, and π. Repeat for second half: 5π/4, 3π/2, 7π/4, and 2π.

12. Graphing Sine/Cosine Curves
Sketch the graph of y = 3 sin 2x in the interval 0 ≤ x ≤ 2π.
max. = 3, min. = -3; period is π
2. Plot points & sketch
y = 3 sin 2x 3 π 2π -3

13. Sketch, on the same set of axes, the graphs of
Model Problem
Sketch, on the same set of axes, the graphs of
y = 2 cos x and y = sin 1/2 x in the interval 0 ≤ x ≤ 2π.
1) Determine the amplitudes & periods
y = 2 cos x
y = sin 1/2 x
a = 2, b = 1
a = 1, b = 1/2
amplitude = | a | = 2
max. = 2 min. = -2
amplitude = | a | = 1
max. = 1 min. = -1
= 2π
|b|
period
|1|
divide the period 2π,
into 4 equal intervals: π/2, π, 3π/2, and 2π.
= 2π
|b|
period
|1/2|
= 4π
divide the period 4π,
into 4 equal intervals: π, 2π, 3π, and 4π.

14. Sketch, on the same set of axes, the graphs of
Model Problem (con’t)
Sketch, on the same set of axes, the graphs of
y = 2 cos x and y = sin 1/2 x in the interval 0 ≤ x ≤ 2π.
y = 2 cos x
max. = 2 min. = -2
period = 2π
y = sin 1/2x
max. = 1 min. = -1
period = 4π
2. Plot points & sketch
y = 2 cos x 2 1
y = sin 1/2x
2 cos x = sin 1/2x π 2π -1 -2

15. Model Problems
The amplitude of y = 2 sin 2x is
1)  ) ) ) 4
What is the range of the function 3 sin x?
1) y > ) y < ) -1 < y < ) -3 < y < 3
What is the minimum value of f() in the equation f() = 3 sin 4
1) ) ) ) -4
What is the period of sin 2x?
1) 4 ) )  4) 4

16. Which is the equation of the graph shown?
Model Problems 2
Which is the equation of the graph shown?
y = 2 sin ½x ) y = 2 cos ½x
3) y = ½ sin 2x 4) y = ½ cos 2x

17. Which is the equation of the graph shown?
Regents Prep 2
Which is the equation of the graph shown?
y = 2 sin ½x ) y = 2 sin 2x
3) y = ½ cos 2x 4) y = 2 cos 2x

18. Model Question
Which function has a period of 4 and an amplitude of 8?
y = -8 sin 8x ) y = -8 sin ½ x
3) y = 8 sin 2x 4) y = 4 sin 8x
The period of a sine function is 300 and its amplitude is 3. Write the function in y = a sin bx form.
y = 3 sin 12x