chapter 6 test review

Post on 22-Feb-2016

43 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

Chapter 6 Test Review. State the values of θ for which each equation is true:. 1.) sin θ = -12.) sec θ = -13.) tan θ = 0. 270° + 360°k. 180° + 360°k. 180°k. 4 .) Sin θ = -15.) Sec θ = -16.) Tan θ = 0. 180°. 0°. -90°. - PowerPoint PPT Presentation

TRANSCRIPT

Chapter 6 Test Review

State the values of θ for which each equation is true:1.) sin θ = -1 2.) sec θ = -1 3.) tan θ = 0

270° + 360°k 180° + 360°k 180°k

4.) Sin θ = -1 5.) Sec θ = -1 6.) Tan θ = 0

-90°180° 0°

State the amplitude, period, and phase shift of each function.

1. y = -2sin θ 2. y = 10sec θ 3. y = -3sin4θ

4. 5. y = 2.5cos(θ + 180°) 6.

A = 2P = 360°PS = 0°

A = 10P = 360°PS = 0°

A = 3P = 90° or π/2PS = 0°

A = 0.5P = 360° or 2πPS = 60° π/3 RIGHT

A = 2.5P = 360° or 2πPS = 180° π LEFT

A = 1.5P = 90° or π/2PS = π/16 RIGHT

Write an equation of the cosine function with amplitude, period, and phase shift given.

1. A = 0.75, P = 360°, PS = 30°

2. A = 4, P = 3°, PS = -30°

y = ±0.75cos(θ – 30°)

y = ±4cos(120θ + 3600°)

Graph: -360° ≤ x ≤ 360°, scale 45°1. y = 2cos (2x – 45°) 2. y = 2sin x + cos x

X 2sinx Cosx SUM0 0 1 190 2 0 2180 0 -1 -1270 -2 0 -2360 0 1 1

Find the values of x (0°≤x≤360°) that satisfy each equation.

1. x = arccos 1 2. arccos = x 3. arcsin ½ = x

4. sin-1 (-1) = x 5. sin-1 = x 6. cot-1 1 = x

cos x = 1

0°, 360°

cos x =

45°, 315°

sin x = ½

30°, 150°

sin x = -1

270°

sin x =

45°, 135°

cot x = 1

45°, 225°

Evaluate. Assume all angles are in quadrant I

1. cos (cos-1 ½) 2. sin (cos-1 ½) 3. cos (sin-1 ½)

4.

1/2 √3/2 √3/2

tan (45° - 45°) = tan 0° = 0

Evaluate.1. 2.

State the domain and range of each function:

1. y = Cos x 2. y = Sin x 3. y = Tan x

4. y = Arccos x 5. y = Sin-1 x 6. y = Arctan x

Domain: 0° ≤ x ≤ 180°Range: -1 ≤ y ≤ 1

Domain: -90° ≤ x ≤ 90°Range: -1 ≤ y ≤ 1

Domain: -90° < x < 90°Range: all reals

Domain: all realsRange: -90° < y < 90°

Domain: -1 ≤ x ≤ 1Range: -90° ≤ y ≤ 90°

Domain: -1 ≤ x ≤ 1Range: 0° ≤ y ≤ 180°

Graph y = Arccos x

Graph y = Arcsin x

Graph the inverse of: y = Sin (x + 90°)

Graph the inverse of: y = Arctan x + π/4

Determine a counterexample for the following statement:

1. Cos-1 x = Cos-1 (-x) 2. Sin-1 x = -Sin-1 x

x = 1 x = 1

3. 4.

x = π/2 or 90° x = 0°

Find the inverse of each function:

1.) y = Cos (x + π) 2.) y = Sin x

3.) y = Sin θ + π/2 4.) y = Sin (x + π/2)

Determine a value for x that would NOT produce a counterexample to the

equation:

x = -1

Write an equation with a phase shift 0 to represent a simple harmonic motion under each set of circumstances.

1.) Initial pos. 12, amplitude 12, period 8 2.) Initial pos. 0, amplitude 2, period 8π

3.) Initial pos. -24, amplitude 24, period 6

The paddle wheel of a boat measures 16 feet in diameter and is revolving at a rate of 20 rpm. If the lowest point of the wheel is 1 foot under water, write an equation in terms of cosine to describe the height of the initial point after “t” seconds.

State the amplitude, period, frequency, and phase shift for the function:

A = 0.4

P =

PS =

Write an equation with phase shift 0 to represent simple harmonic motion with initial position 0, amplitude 5, and period 3

Initial position 0 means it is a sine function.

Period = 2π / k = 3

k = 2π/3

Write an equation with phase shift 0 to represent simple harmonic motion with initial position -12, amplitude 12, and period ½

Initial position -12 means it is a cosine function.

Period = 2π / k = ½

k = 4π

The paddle wheel of a boat measures 16 feet in diameter and is revolving at a rate of 20 rpm. If the lowest point of the wheel is 1 foot under water, write an equation in terms of cosine to describe the height of the initial point after t seconds.

Evaluate. 1. sin (Sin-1 ½) 2. tan (Arccos ½) 3. Cos (tan π/4)

4.

1/2 √3 0

tan (120° + 60°) = cos 180° = -1

top related