# How do you find the amplitude and period of a function y = sin(pi/2Ө)?

##### 1 Answer

Amplitude$= 1$

Period $= 4$

#### Explanation:

from the given $y = \sin \left(\frac{\pi}{2} \cdot \theta\right)$

it should be taken into account that the above is the same as

$y = 1 \cdot \sin \left(\frac{\pi}{2} \cdot \theta\right)$

Take the absolute value of any number you see as numerical coefficient of sine or cosine expression and you get the Amplitude.

Amplitude $= \left\mid 1 \right\mid = 1$

The Period can be obtain using $P = \frac{2 \pi}{a}$ where $a$ is the number beside $\theta$

so that $P = \frac{2 \pi}{\left\mid \frac{\pi}{2} \right\mid} = 4$

Learn from the following examples,

$y = - 7 \cdot \sin 3 \theta$
Amplitude$= \left\mid - \right\mid 7 = 7$ and Period$= \frac{2 \pi}{\left\mid 3 \right\mid} = \frac{2 \pi}{3}$

$y = - \frac{3}{4} \cdot \cos \left(\frac{\pi \theta}{6}\right)$

Amplitude$= \left\mid - \frac{3}{4} \right\mid = \frac{3}{4}$and Period$= \frac{2 \pi}{\left\mid \frac{\pi}{6} \right\mid} = 12$

$y = 12 \cdot \sin \left(\theta\right)$

Amplitude$= \left\mid 12 \right\mid = 12$and Period$= \frac{2 \pi}{\left\mid 1 \right\mid} = 2 \pi$

I hope the additional examples help..

Have a nice day from the Philippines ....