# How do you find the period and amplitude of y=1/4sin(2pix)?

Jul 15, 2018

The period is $T = 1$. The amplitude is $= \frac{1}{4}$

#### Explanation:

The period $T$ of a periodic function $f \left(x\right)$ is defined by

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

Here,

$f \left(x\right) = \frac{1}{4} \sin \left(2 \pi x\right)$............................$\left(1\right)$

Therefore,

$f \left(x + T\right) = \frac{1}{4} \sin 2 \pi \left(x + T\right)$

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

$= \frac{1}{4} \sin \left(2 \pi x\right) \cos \left(2 \pi T\right) + \frac{1}{4} \cos \left(2 \pi x\right) \sin \left(2 \pi T\right)$...........................$\left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$

$\left\{\begin{matrix}\cos 2 \pi T = 1 \\ \sin 2 \pi T = 0\end{matrix}\right.$

$\implies$, $2 \pi T = 2 \pi$

The period is $T = 1$

For the sine function

$- 1 \le \sin x \le 1$

$- \frac{1}{4} \le \frac{1}{4} \sin x \le \frac{1}{4}$

$- \frac{1}{4} \le \frac{1}{4} \sin 2 \pi x \le \frac{1}{4}$

The amplitude is $= \frac{1}{4}$

graph{1/4sin(2pix) [-0.587, 2.113, -0.696, 0.654]}