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Determine the amplitude, period, and phase shift of the following trigonometric equation.

[tex]\[ y = 2 \sin(x) \][/tex]

Answer:

Amplitude: [tex]\(\square\)[/tex]

Phase Shift:
- no phase shift
- shifted to the right
- shifted to the left


Sagot :

To determine the amplitude, period, and phase shift of the trigonometric equation [tex]\( y = 2 \sin(x) \)[/tex], we need to analyze the standard form of the sine function, which is:

[tex]\[ y = A \sin(Bx + C) \][/tex]

Here, [tex]\( A \)[/tex] represents the amplitude, [tex]\( B \)[/tex] affects the period, and [tex]\( C \)[/tex] influences the phase shift.

1. Amplitude [tex]\( (A) \)[/tex]:

The amplitude is the coefficient of the sine function, which affects the height of the wave. For the given equation [tex]\( y = 2 \sin(x) \)[/tex], the coefficient [tex]\( A \)[/tex] is 2. Therefore, the amplitude is:

[tex]\[ \text{Amplitude} = 2 \][/tex]

2. Period:

The period of the sine function is determined by the coefficient [tex]\( B \)[/tex] inside the argument of the sine function. The standard period of [tex]\( \sin(x) \)[/tex] is [tex]\( 2\pi \)[/tex]. When the argument is [tex]\( Bx \)[/tex], the period changes to [tex]\( \frac{2\pi}{B} \)[/tex]. In the given equation, since [tex]\( B = 1 \)[/tex], the period remains:

[tex]\[ \text{Period} = 2\pi \][/tex]

3. Phase Shift:

The phase shift is determined by the term [tex]\( C \)[/tex] inside the sine function. The phase shift is calculated as [tex]\( -\frac{C}{B} \)[/tex]. In this case, since the given equation does not include any additional term inside the sine function (i.e., [tex]\( C = 0 \)[/tex]), there is no horizontal shift. Therefore, the phase shift is:

[tex]\[ \text{Phase Shift} = 0 \][/tex]
This indicates there is no phase shift for the given function.

Hence, summarizing the results:
- Amplitude: [tex]\( 2 \)[/tex]
- Period: [tex]\( 2\pi \)[/tex]
- Phase Shift: No phase shift