Answer :
Sure, I'd be happy to walk you through the detailed solution for each part of the given question.
### Part (a): Amplitude
The amplitude of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the absolute value of the coefficient [tex]\( a \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The coefficient of [tex]\(\sin\)[/tex] is [tex]\(4\)[/tex].
- Therefore, the amplitude is [tex]\(4\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The coefficient of [tex]\(\cos\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex].
- Therefore, the amplitude is the absolute value of [tex]\(-\frac{1}{2}\)[/tex], which is [tex]\(\frac{1}{2}\)[/tex].
### Part (b): Period
The period of a sine or cosine function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the formula [tex]\( \frac{2π}{|b|} \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The coefficient inside the sine function (b) is [tex]\(\frac{1}{3}\)[/tex].
- The period is [tex]\( \frac{2π}{|\frac{1}{3}|} \)[/tex].
- Simplifying this, we get [tex]\( 2π \times 3 = 6π \)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The coefficient inside the cosine function (b) is [tex]\(4\)[/tex].
- The period is [tex]\( \frac{2π}{|4|} \)[/tex].
- Simplifying this, we get [tex]\( \frac{2π}{4} = \frac{π}{2} \)[/tex].
### Part (c): Phase Shift
The phase shift of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by solving [tex]\( bx + c = 0 \)[/tex] for [tex]\( x \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- Set the inside of the sine function to zero: [tex]\( \frac{1}{3} (x + 30°) = 0 \)[/tex].
- Solve for [tex]\( x \)[/tex]: [tex]\( x + 30° = 0 \)[/tex] [tex]\(\Rightarrow x = -30°\)[/tex].
- Therefore, the phase shift is [tex]\(-30°\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- Set the inside of the cosine function to zero: [tex]\( 4 (x + 135°) = 0 \)[/tex].
- Solve for [tex]\( x \)[/tex]: [tex]\( x + 135° = 0 \)[/tex] [tex]\(\Rightarrow x = -135°\)[/tex].
- Therefore, the phase shift is [tex]\(-135°\)[/tex].
### Part (d): Vertical Shift
The vertical shift of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the constant [tex]\( d \)[/tex] at the end of the expression.
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The vertical shift is [tex]\(-1\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The vertical shift is [tex]\(2\)[/tex].
### Summary of Results
a) Amplitude:
- Function (i): [tex]\(4\)[/tex]
- Function (ii): [tex]\(\frac{1}{2}\)[/tex]
b) Period:
- Function (i): [tex]\(6π\)[/tex] (approximately [tex]\(18.85\)[/tex])
- Function (ii): [tex]\(\frac{π}{2}\)[/tex] (approximately [tex]\(1.57\)[/tex])
c) Phase Shift:
- Function (i): [tex]\(-30°\)[/tex]
- Function (ii): [tex]\(-135°\)[/tex]
d) Vertical Shift:
- Function (i): [tex]\(-1\)[/tex]
- Function (ii): [tex]\(2\)[/tex]
### Part (a): Amplitude
The amplitude of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the absolute value of the coefficient [tex]\( a \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The coefficient of [tex]\(\sin\)[/tex] is [tex]\(4\)[/tex].
- Therefore, the amplitude is [tex]\(4\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The coefficient of [tex]\(\cos\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex].
- Therefore, the amplitude is the absolute value of [tex]\(-\frac{1}{2}\)[/tex], which is [tex]\(\frac{1}{2}\)[/tex].
### Part (b): Period
The period of a sine or cosine function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the formula [tex]\( \frac{2π}{|b|} \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The coefficient inside the sine function (b) is [tex]\(\frac{1}{3}\)[/tex].
- The period is [tex]\( \frac{2π}{|\frac{1}{3}|} \)[/tex].
- Simplifying this, we get [tex]\( 2π \times 3 = 6π \)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The coefficient inside the cosine function (b) is [tex]\(4\)[/tex].
- The period is [tex]\( \frac{2π}{|4|} \)[/tex].
- Simplifying this, we get [tex]\( \frac{2π}{4} = \frac{π}{2} \)[/tex].
### Part (c): Phase Shift
The phase shift of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by solving [tex]\( bx + c = 0 \)[/tex] for [tex]\( x \)[/tex].
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- Set the inside of the sine function to zero: [tex]\( \frac{1}{3} (x + 30°) = 0 \)[/tex].
- Solve for [tex]\( x \)[/tex]: [tex]\( x + 30° = 0 \)[/tex] [tex]\(\Rightarrow x = -30°\)[/tex].
- Therefore, the phase shift is [tex]\(-30°\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- Set the inside of the cosine function to zero: [tex]\( 4 (x + 135°) = 0 \)[/tex].
- Solve for [tex]\( x \)[/tex]: [tex]\( x + 135° = 0 \)[/tex] [tex]\(\Rightarrow x = -135°\)[/tex].
- Therefore, the phase shift is [tex]\(-135°\)[/tex].
### Part (d): Vertical Shift
The vertical shift of a trigonometric function [tex]\( y = a \sin(bx + c) + d \)[/tex] or [tex]\( y = a \cos(bx + c) + d \)[/tex] is given by the constant [tex]\( d \)[/tex] at the end of the expression.
1. For the function [tex]\( y = 4 \sin \left[ \frac{1}{3} (x + 30°) \right] - 1 \)[/tex]:
- The vertical shift is [tex]\(-1\)[/tex].
2. For the function [tex]\( y = -\frac{1}{2} \cos \left[ 4 (x + 135°) \right] + 2 \)[/tex]:
- The vertical shift is [tex]\(2\)[/tex].
### Summary of Results
a) Amplitude:
- Function (i): [tex]\(4\)[/tex]
- Function (ii): [tex]\(\frac{1}{2}\)[/tex]
b) Period:
- Function (i): [tex]\(6π\)[/tex] (approximately [tex]\(18.85\)[/tex])
- Function (ii): [tex]\(\frac{π}{2}\)[/tex] (approximately [tex]\(1.57\)[/tex])
c) Phase Shift:
- Function (i): [tex]\(-30°\)[/tex]
- Function (ii): [tex]\(-135°\)[/tex]
d) Vertical Shift:
- Function (i): [tex]\(-1\)[/tex]
- Function (ii): [tex]\(2\)[/tex]
