How does the number in front effect the graph? Graph on GUT y1 = sinx y2 = 2 sinx y3 = 3 sinx y4 = ½ sinx How does the number in front effect the graph? Trigonometry

Amplitude and period of sine and cosine functions 6.4 Amplitude and period of sine and cosine functions Trigonometry

Amplitude Amplitude is defined to be ½ the distance between the lowest and highest points on the graph. The “amplitude” of y = A sin x is |A| Because it is defined to be a distance amplitude is always positive. Trigonometry

Amplitude examples: Y = 3 sin x Y = -2 sinx Y = 1/3 cos x What does the negative do to the graph? Reflects over x axis. Trigonometry

How does the number in front of x effect the graph? Graph on GUT y1 = sin x y2 = sin 2x y3 = sin 3x y4 = sin ½x How does the number in front of x effect the graph? Trigonometry

Period of a function The period is the distance it takes for a graph to “do its thing.” Period of a y = A sin x or y = A cos x is 2π. The period of y = A sin kx or y = A cos kx is 2π/k. If the p < 2π then graph is squished horizontally. If the p > 2π then the graph is stretched horizontally. Trigonometry

State the amplitude and period for each. f(x) = 4 cos x f(x) = -2 sin ½ x f(x) = 1/3 cos 2x A = 4 p = 2π A = 2 p = 4π A = 1/3 p = π Trigonometry

Basic Forms Y = ± A sin (kx) Y = ± A cos (kx) Trigonometry

Graph each. Y = -3 sin (x/4), -4π < x < 8π A = |-3| A = 3 P = 2π/k P = 2π/(1/4) P = 8π Trigonometry

Graph each. 2. Y = -2 cos (x/2), -4π < x < 8π A = |-2| A = 2 P = 2π/k P = 2π/(1/2) P = 4π Trigonometry

Graph each. 3. Y = 1/2 sin (4x), -π < x < π A = |1/2| A = 1/2 Trigonometry

Graph. 4. y = 4 sin x, -2π < x < 2π A = |4| A = 4 P = 2π/k Trigonometry

Graph. 5. y = 3 cos (x/4), -π < x < π A = |3| A = 3 P = 2π/k Trigonometry

Graph 6. Y = 1/3 cos (4x), -π < x < π A = |1/3| A = 1/3 P = 2π/k Trigonometry

Find the amplitude and period for each function. 1. A piano tuner strikes a tuning fork for note A above middle C and sets in motion vibrations can by modeled by the equation y = 0.001 sin 880π t. 2. A buoy that bobs up and down in the waves can be modeled by y= 1.75 cos π/3 t. 3. A pendulum can be modeled by the function d= 4 cos 8π t, where d is the horizontal displacement and t is time. Trigonometry

Y = ± A cos (kx) |A| = 9.8 A = ±9.8 p = 6π 2π/k = 6π 2π = 6πk ⅓ = k Write the equation of the cosine function with amplitude 9.8 and period 6π. Y = ± A cos (kx) |A| = 9.8 A = ±9.8 p = 6π 2π/k = 6π 2π = 6πk ⅓ = k Y = ± 9.8 cos ⅓ x or y = ± 9.8 cos x/3 Trigonometry

Y = ± A sin (kx) |A| = 4.1 A = ±4.1 p = π/2 2π/k = π/2 4π = πk 4 = k Write the equation of the sine function with amplitude 4.1 and period π/2. Y = ± A sin (kx) |A| = 4.1 A = ±4.1 p = π/2 2π/k = π/2 4π = πk 4 = k Y = ± 4.1 sin 4x Trigonometry

Y = ± A cos (kx) |A| = 2 A = ±2 p = π/2 2π/k = π/2 4π = πk 4 = k Write the equation of the cosine function with amplitude 2 and period π/2. Y = ± A cos (kx) |A| = 2 A = ±2 p = π/2 2π/k = π/2 4π = πk 4 = k Y = ± 2 cos 4x Trigonometry

Y = ± A sin (kx) |A| = 0.5 A = ±0.5 p = 0.2π 2π/k = .2π 2π = .2πk Write the equation of the sine function with amplitude 0.5 and period 0.2π. Y = ± A sin (kx) |A| = 0.5 A = ±0.5 p = 0.2π 2π/k = .2π 2π = .2πk 10 = k Y = ± 0.5 sin 10x Trigonometry

Y = ± A cos (kx) |A| = 1/5 A = ± 1/5 p = 2/5 π 2π/k = 2π/5 10π = 2πk Write the equation of the cosine function with amplitude 1/5 and period 2/5 π. Y = ± A cos (kx) |A| = 1/5 A = ± 1/5 p = 2/5 π 2π/k = 2π/5 10π = 2πk 5 = k Y = ± 1/5 cos 5x Trigonometry

A = -3.5/2 (negative because it is on its way down) 2π/k = 14 2π = 14k A signal buoy bobs up and down. From the highest point to the lowest point, the buoy moves a distance of 3.5 feet. It moves from its highest point down to its lowest point and back to its highest point every 14 seconds. Write an equation of the motion for the buoy assuming that it is at its equilibrium point at t = 0 and the buoy is on its way down at that time. Y = ± A sin kt A = -3.5/2 (negative because it is on its way down) 2π/k = 14 2π = 14k π/7 = k y = -1.75 sin π/7 t Trigonometry

Determine the height of the buoy at 8 seconds and at 17 seconds A signal buoy bobs up and down. From the highest point to the lowest point, the buoy moves a distance of 3.5 feet. It moves from its highest point down to its lowest point and back to its highest point every 14 seconds. Determine the height of the buoy at 8 seconds and at 17 seconds y = -1.75 sin π/7 t y = -1.76 sin π/7 (8) y ≈ 0.75 After 8 seconds, the buoy is about .8 feet above the equilibrium point. y = -1.76 sin π/7 (17) y ≈ -1.71 After 17 seconds, the buoy is about 1.71 feet below the equilibrium point. Trigonometry

A buoy bobs up and down. The distance between the highest and lowest point is 3 feet. It moves from its highest point down to its lowest point and back to its highest point every 8 seconds. Find the equation of the motion for the buoy assuming that it is at its equilibrium point at t=0 and the buoy is on its way down up at that time. Y = ± A sin kt A = 3/2 (positive because it is on its way up) 2π/k = 8 2π = 8k π/4 = k y = 1.5 sin π/4 t Trigonometry

A buoy bobs up and down. The distance between the highest and lowest point is 3 feet. It moves from its highest point down to its lowest point and back to its highest point every 8 seconds. (b) Determine the height of the buoy at 3 seconds. y = 1.5 sin π/4 t y = 1.5 sin π/4 (3) y = 3.18 feet (c) Determine the height of the buoy at 12 seconds. y = 1.5 sin π/4 t y = 1.5 sin π/4 (12) y = 12.73 feet Trigonometry

A circuit has an alternating voltage of 100 volts that peaks every 0 A circuit has an alternating voltage of 100 volts that peaks every 0.5 seconds. Write a sin function for the voltage as a function of time t (in seconds). Trigonometry

Frequency: the number of cycles per unit of time Frequency = 1/period Period = 1/frequency hertz is a unit of frequency, One hertz = one cycle per second Trigonometry

Write an equation of the sine function that represents the initial behavior of the vibrations of the note G above middle C having amplitude of 0.015 and frequency of 392 hertz. |A| = 0.015 A = ± 0.015 P = 1/frequency P = 1/392 P = 2π/k 1/392 = 2π/k K = 784π y = ±A sin kx A = ± 0.015 sin 784πt Trigonometry

The Sears Building in Chicago sways back and forth at a vibration frequency of about 0.1 Hz. On average it sways 6 inches from true center. Write an equation of the sine function that represents this behavior. |A| = 6 A = ± 6 P = 1/frequency P = 1/.1 P = 10 P = 2π/k 10 = 2π/k 10K = 2π k = π/5 y = ±A sin kx A = ± 6 sin π/5 t Trigonometry

Exit Pass State the amplitude and period for f(x) = -2 sin (x/3). Graph y = 2 cos (4x) –π < x < π Graph: y = -3 sin (x/2) -2π < x < 6π Trigonometry

Even Answers Trigonometry