Transcript PPT Chapter 6 Review-0x

Chapter 6 Test Review

State the values of θ for which each
equation is true:
1.) sin θ = -1
270° + 360°k
4.) Sin θ = -1
-90°
2.) sec θ = -1
180° + 360°k
5.) Sec θ = -1
180°
3.) tan θ = 0
180°k
6.) Tan θ = 0
0°
State the amplitude, period, and phase
shift of each function.
1. y = -2sin θ
2. y = 10sec θ
A=2
P = 360°
PS = 0°
A = 10
P = 360°
PS = 0°
4. y = 0.5sinæçq - p ö÷
5. y = 2.5cos(θ + 180°)
è
3ø
A = 2.5
A = 0.5
P = 360° or 2π
P = 360° or 2π
PS = 60° π/3 RIGHT PS = 180° π LEFT
3. y = -3sin4θ
A=3
P = 90° or π/2
PS = 0°
6. y = -1.5sinæç 4q - p ö÷
è
4ø
A = 1.5
P = 90° or π/2
PS = 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°
y = ±0.75cos(θ – 30°)
2. A = 4, P = 3°, PS = -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
SUM
0
0
1
1
90
2
0
2
180
0
-1
-1
270
-2
0
-2
360
0
1
1
Find the values of x (0°≤x≤360°) that
satisfy each equation.
1. x = arccos 1
2. arccos
cos x = 1
cos x =
0°, 360°
45°, 315°
4. sin-1 (-1) = x
5. sin-1
2= x
2
2
2
3. arcsin ½ = x
sin x = ½
30°, 150°
2=x
2
6. cot-1 1 = x
2
2
cot x = 1
sin x = -1
sin x =
270°
45°, 135°
45°, 225°
Evaluate. Assume all angles are in
quadrant I
1. cos (cos-1 ½)
1/2
4.
2. sin (cos-1 ½)
√3/2
æ
ö
-1 2
-1 2
tançsin
- cos
÷
2
2
è
ø
tan (45° - 45°) = tan 0° = 0
3. cos (sin-1 ½)
√3/2
Evaluate.
1.
ép
ù
-1
sinê + Cos (0) ú
ë3
û
ép p ù
sin ê + ú
ë3 2û
é 5p ù 1
sin ê ú =
ë6 û 2
2.
é 3p
ù
-1
tanê + Sin (0)ú
ë4
û
é 3p
ù
tanê + 0 ú
ë4
û
é 3p ù
tanê ú = -1
ë4 û
State the domain and range of each
function:
1. y = Cos x
Domain: 0° ≤ x ≤ 180°
Range: -1 ≤ y ≤ 1
4. y = Arccos x
Domain: -1 ≤ x ≤ 1
Range: 0° ≤ y ≤ 180°
2. y = Sin x
Domain: -90° ≤ x ≤ 90°
Range: -1 ≤ y ≤ 1
5. y = Sin-1 x
Domain: -1 ≤ x ≤ 1
Range: -90° ≤ y ≤ 90°
3. y = Tan x
Domain: -90° < x < 90°
Range: all reals
6. y = Arctan x
Domain: all reals
Range: -90° < y < 90°
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. Cos-1 x =
1
Cos(x)
x = π/2 or 90°
4.
Tan -1 x =
x = 0°
1
Tan(x)
Find the inverse of each function:
1.) y = Cos (x + π)
2.) y = Sin x
x = Cos(y + p )
x = Sin(y)
Arc cos x = y + p
Arc sin x = y
Arc cos x - p = y
3.) y = Sin θ + π/2
x-
2
p
p
2
x = Sin(y + )
2
= Sin(y)
Arc sin x = y +
x = Sin(y) +
p
4.) y = Sin (x + π/2)
æ
pö
Arc sinç x - ÷ = y
è
2ø
Arc sin x -
p
2
p
2
=y
Determine a value for x that would NOT produce a
counterexample to the equation:
-1
-1
Sin x = Tan x -
p
4
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
p
t
y = ±2sin
4
y = 12cos t
4
3.) Initial pos. -24, amplitude 24, period 6
p
y = -24 cos t
3
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.
æ 2p ö
h = -8cosç t ÷ + 7
è 3 ø
State the amplitude, period, frequency,
and phase shift for the function:
æ
pö
y = -0.4cos ç 20p t + ÷
è
2ø
A = 0.4
2p
1
P = 20p = 10
-p
2 = - p × 1 = 1 left
PS =
20p
2 20p 40
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
æ 2p ö
y = 5sin ç t ÷
è 3 ø
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
y =12sin ( 4p t )
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)
1/2
4.
2. tan (Arccos ½)
√3
æ
3ö
-1 -1
-1
cos ç Cos
+ Sin
÷
2
2
è
ø
tan (120° + 60°) = cos 180° = -1
3. Cos (tan π/4)
0