# How do you find the amplitude, period, and phase shift of 4cos(3theta + 3/2pi) + 2?

Jun 4, 2015

First, the range of the cosinus function is [-1;1]
$\rightarrow$ therefore the range of $4 \cos \left(X\right)$ is [-4;4]
$\rightarrow$ and the range of $4 \cos \left(X\right) + 2$ is [-2;6]

Second, the period $P$ of the cosinus function is defined as: $\cos \left(X\right) = \cos \left(X + P\right)$ $\rightarrow P = 2 \pi$.
$\rightarrow$ therefore:

$\left(3 {\theta}_{2} + \frac{3}{2} \pi\right) - \left(3 {\theta}_{1} + \frac{3}{2} \pi\right) = 3 \left({\theta}_{2} - {\theta}_{1}\right) = 2 \pi$

$\rightarrow$ the period of $4 \cos \left(3 \theta + \frac{3}{2} \pi\right) + 2$ is $\frac{2}{3} \pi$

Third, $\cos \left(X\right) = 1$ if $X = 0$
$\rightarrow$ here $X = 3 \left(\theta + \frac{\pi}{2}\right)$
$\rightarrow$ therefore $X = 0$ if $\theta = - \frac{\pi}{2}$
$\rightarrow$ therefore the phase shift is $- \frac{\pi}{2}$