# How do you find the amplitude and period of y=3cos(2/3theta)?

Nov 1, 2016

For $y = a \cos \left(b \theta\right) = 3 \cos \left(\frac{3}{2} \theta\right)$
$A m p l i t u \mathrm{de} = a = 3$
Period = (2π)/|b| = (2pi)/(2/3) = (6pi)/2 = 3pi

#### Explanation:

For $a m p l i t u \mathrm{de}$, we need the factor $A$,
so for $y = 3 \cos \left(\frac{3}{2} \theta\right)$, that would be 3.

For $p e r i o d$, we must plug-in $b$ to (2π)/|b| and simplify.
$b = \frac{3}{2}$, so (2π)/|b| = (2pi)/(2/3). To simplify this, we must first multiply the lowest denominator by the numerator $\frac{\left(2 \pi\right) \cdot 3}{\left(\frac{2}{3}\right) \cdot 3}$ getting $\frac{6 \pi}{2}$, which we can then simplify dividing by 2 $\frac{\frac{6 \pi}{2}}{\frac{2}{2}}$ getting $3 \pi$.