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To rewrite the equation [tex]\( y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) \)[/tex] in the form [tex]\( y(t) = A \sin(\omega t + \phi) \)[/tex], follow these steps:
1. Identify the coefficients of sine and cosine:
- [tex]\( a = 2 \)[/tex] (coefficient of [tex]\( \sin(4\pi t) \)[/tex])
- [tex]\( b = 5 \)[/tex] (coefficient of [tex]\( \cos(4\pi t) \)[/tex])
2. Calculate the amplitude [tex]\( A \)[/tex]:
The amplitude [tex]\( A \)[/tex] can be determined using the Pythagorean theorem applied to the coefficients of the trigonometric components:
[tex]\[ A = \sqrt{a^2 + b^2} \][/tex]
Plugging in the values,
[tex]\[ A = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385 \][/tex]
3. Determine the angular frequency [tex]\( \omega \)[/tex]:
In the original equation, the angular frequency [tex]\( \omega \)[/tex] is associated with both trigonometric functions. Given the term [tex]\( 4\pi t \)[/tex], we have:
[tex]\[ \omega = 4\pi \][/tex]
4. Calculate the phase angle [tex]\( \phi \)[/tex]:
The phase angle [tex]\( \phi \)[/tex] can be found using the arctangent function, specifically the four-quadrant inverse tangent function [tex]\( \text{atan2} \)[/tex], which accounts for the correct quadrant of the angle:
[tex]\[ \phi = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{5}{2}\right) \][/tex]
Evaluating this:
[tex]\[ \phi \approx 1.190 \text{ radians} \][/tex]
Therefore, the equation [tex]\( y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) \)[/tex] can be rewritten in the desired form [tex]\( y(t) = A \sin(\omega t + \phi) \)[/tex]:
[tex]\[ y(t) = 5.385 \sin(12.566 t + 1.190) \][/tex]
Here, we have:
- Amplitude [tex]\( A = 5.385 \)[/tex]
- Angular frequency [tex]\( \omega = 12.566 \)[/tex]
- Phase angle [tex]\( \phi = 1.190 \)[/tex] radians.
1. Identify the coefficients of sine and cosine:
- [tex]\( a = 2 \)[/tex] (coefficient of [tex]\( \sin(4\pi t) \)[/tex])
- [tex]\( b = 5 \)[/tex] (coefficient of [tex]\( \cos(4\pi t) \)[/tex])
2. Calculate the amplitude [tex]\( A \)[/tex]:
The amplitude [tex]\( A \)[/tex] can be determined using the Pythagorean theorem applied to the coefficients of the trigonometric components:
[tex]\[ A = \sqrt{a^2 + b^2} \][/tex]
Plugging in the values,
[tex]\[ A = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385 \][/tex]
3. Determine the angular frequency [tex]\( \omega \)[/tex]:
In the original equation, the angular frequency [tex]\( \omega \)[/tex] is associated with both trigonometric functions. Given the term [tex]\( 4\pi t \)[/tex], we have:
[tex]\[ \omega = 4\pi \][/tex]
4. Calculate the phase angle [tex]\( \phi \)[/tex]:
The phase angle [tex]\( \phi \)[/tex] can be found using the arctangent function, specifically the four-quadrant inverse tangent function [tex]\( \text{atan2} \)[/tex], which accounts for the correct quadrant of the angle:
[tex]\[ \phi = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{5}{2}\right) \][/tex]
Evaluating this:
[tex]\[ \phi \approx 1.190 \text{ radians} \][/tex]
Therefore, the equation [tex]\( y(t) = 2 \sin(4\pi t) + 5 \cos(4\pi t) \)[/tex] can be rewritten in the desired form [tex]\( y(t) = A \sin(\omega t + \phi) \)[/tex]:
[tex]\[ y(t) = 5.385 \sin(12.566 t + 1.190) \][/tex]
Here, we have:
- Amplitude [tex]\( A = 5.385 \)[/tex]
- Angular frequency [tex]\( \omega = 12.566 \)[/tex]
- Phase angle [tex]\( \phi = 1.190 \)[/tex] radians.
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