# How do you find the amplitude, period and phase shift for y=4sin(1/2(theta-pi/2))+5?

Apr 21, 2017

The amplitude is $A = 4$
The period is $= 4 \pi$
The phase shift is $= \frac{\pi}{2}$

#### Explanation:

$y = A \sin \left(B x + C\right) + D$

Amplitude is $A$

Period is (2π)/B

Phase shift is −C/B

Vertical shift is $D$

Here, we have

$y = 4 \sin \left(\frac{1}{2} \left(\theta - \frac{\pi}{2}\right)\right) + 5$

$y = 4 \sin \left(\frac{1}{2} \theta - \frac{\pi}{4}\right) + 5$

The amplitude is $A = 4$

The period is $= \frac{2 \pi}{B} = \left(2 \frac{\pi}{\frac{1}{2}}\right) = 4 \pi$

The phase shift is $= \frac{\frac{\pi}{4}}{\frac{1}{2}} = \frac{\pi}{2}$

graph{4sin(1/2(x-pi/2))+5 [-11.71, 13.61, -1.49, 11.17]}