# What is the amplitude, period and the phase shift of y = 2 sin (1/4 x)?

May 26, 2018

The amplitude is $= 2$. The period is $= 8 \pi$ and the phase shift is $= 0$

#### Explanation:

We need

$\sin \left(a + b\right) = \sin a \cos b + \sin b \cos a$

The period of a periodic function is $T$ iif

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

Here,

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

Therefore,

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

where the period is $= T$

So,

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

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

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

Then,

$\left\{\begin{matrix}\cos \left(\frac{1}{4} T\right) = 1 \\ \sin \left(\frac{1}{4} T\right) = 0\end{matrix}\right.$

$\iff$, $\frac{1}{4} T = 2 \pi$

$\iff$, $T = 8 \pi$

As

$- 1 \le \sin t \le 1$

Therefore,

$- 1 \le \sin \left(\frac{1}{4} x\right) \le 1$

$- 2 \le 2 \sin \left(\frac{1}{4} x\right) \le 2$

The amplitude is $= 2$

The phase shift is $= 0$ as when $x = 0$

$y = 0$

graph{2sin(1/4x) [-6.42, 44.9, -11.46, 14.2]}

May 26, 2018

$2 , 8 \pi , 0$

#### Explanation:

$\text{the standard form of the sine function is}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a \sin \left(b x + c\right) + d} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{amplitude "=|a|," period } = \frac{2 \pi}{b}$

$\text{phase shift "=-c/b" and vertical shift } = d$

$\text{here } a = 2 , b = \frac{1}{4} , c = d = 0$

$\text{amplitude "=|2|=2," period } = \frac{2 \pi}{\frac{1}{4}} = 8 \pi$

$\text{there is no phase shift}$