Determine the period of the function y = - 2 sin(pi(t)/4 + pi/2)

Mathematics
1

determine the period of the function y = - 2 sin(pi(t)/4 + pi/2)

2

Determine the period of the function ( y = -2 \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) )

Answer:
To find the period of the function ( y = -2 \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) ), we need to analyze the argument of the sine function and determine how it repeats.

  1. Identify the Argument of the Sine Function:
    The function is of the form ( y = A \sin(Bt + C) ), where ( A ), ( B ), and ( C ) are constants.

    In this case, ( y = -2 \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) ), we have:

    • ( A = -2 )
    • B = \frac{\pi}{4}
    • C = \frac{\pi}{2}

  2. Determine the Period of the Sine Function:
    The standard form of the sine function is ( y = \sin(\omega t + \phi) ), where ( \omega ) is the angular frequency. The period ( T ) of the sine function ( \sin(\omega t) ) is given by:

    T = \frac{2\pi}{\omega}

    By comparing y = \sin(\omega t + \phi) with y = \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) , we can see that \omega = \frac{\pi}{4}.

  3. Calculate the Period ( T ):
    Substitute \omega = \frac{\pi}{4} into the period formula:

    T = \frac{2\pi}{\frac{\pi}{4}} = \frac{2\pi \times 4}{\pi} = 8

Thus, the period of the function $ y = -2 \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) $s ( 8 ).

Final Answer:
The period of the function y = -2 \sin\left(\frac{\pi t}{4} + \frac{\pi}{2}\right) is \boxed{8}.