# How do you determine the amplitude and period for y = -1/2 cos(pi/3)x?

Jun 20, 2018

The amplitude is $= \frac{1}{2}$ and the period is $= 6$

#### Explanation:

The amplitude is $= \frac{1}{2}$ as $- 1 \le \sin x \le 1$

This a periodic function such that

$f \left(x\right) = f \left(x + T\right)$

Therefore,

$- \frac{1}{2} \cos \left(\frac{\pi}{3} x\right) = - \frac{1}{2} \cos \left(\frac{\pi}{3} \left(x + T\right)\right)$

$= - \frac{1}{2} \cos \left(\frac{\pi}{3} x + \frac{\pi}{3} T\right)$

$= - \frac{1}{2} \left(\cos \left(\frac{\pi}{3} x\right) \cos \left(\frac{\pi}{3} T\right) - \sin \left(\frac{\pi}{3} x\right) \sin \left(\frac{\pi}{3} T\right)\right)$

Therefore,

$\left\{\begin{matrix}\cos \left(\frac{\pi}{3} T\right) = 1 \\ \sin \left(\frac{\pi}{3} T\right) = 0\end{matrix}\right.$

$\iff$, $\frac{\pi}{3} T = 2 \pi$

$\iff$, $T = 6$

The period is $T = 6$

graph{-1/2cos(pi/3x) [-4.05, 11.75, -3.027, 4.875]}