How do you determine the amplitude, period, and phase shift for $y = sin (x - \frac{π}{2})$ $y=sin(x-pi/2)$ ?

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How do you determine the amplitude, period, and phase shift for $y = sin (x - \frac{π}{2})$ $y=sin(x-pi/2)$ ?

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Answer 1 · 2015-04-25T13:06:35

$sin (θ)$ $sin(theta)$ has
an amplitude of $1$ $1$
a period of $2 π$ $2pi$
and a phase shift of $0$ $0$

The only effect of replacing $θ with (x - \frac{π}{2})$ $theta " with " (x-pi/2)$
is to shift the variable to the left by $\frac{π}{2}$ $pi/2$
(e.g.. $sin (x - \frac{π}{2}) with x = \frac{π}{2}$ $sin(x - pi/2) " with " x=pi/2$
is equivalent to $sin (θ) with θ = 0$ $sin(theta) " with " theta=0$ )

So
the amplitude remains $1$ $1$
the period remains $2 π$ $2pi$
and the phase shift is $- \frac{π}{2}$ $-pi/2$

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How do you determine the amplitude, period, and phase shift for $y = sin (x - \frac{π}{2})$ $y=sin(x-pi/2)$ ?

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