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

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

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

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

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.

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.

What about the b? y = cos x y = cos 2x y = cos 1/2x y = cos 4x 1 -1 1 π 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

What about the b? y = sin x y = sin 2x y = sin 1/2x y sin 4x 1 -1 1 -1 π 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

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).

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

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.

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

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.

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

Model Problems The amplitude of y = 2 sin 2x is 1)  2) 2 3) 3 4) 4 What is the range of the function 3 sin x? 1) y > 0 2) y < 0 3) -1 < y < 1 4) -3 < y < 3 What is the minimum value of f() in the equation f() = 3 sin 4 1) -1 2) -2 3) -3 4) -4 What is the period of sin 2x? 1) 4 2) 2 3)  4) 4

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

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

Model Question Which function has a period of 4 and an amplitude of 8? y = -8 sin 8x 2) 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