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

##### 2 Answers

#### Explanation:

The standard form of the

#color(blue)"sine wave function"# is

#color(red)(|bar(ul(color(white)(a/a)color(black)(y=asin(bx+c)+d)color(white)(a/a)|)))# where amplitude = |a| , period

#=(2pi)/b# c is the horizontal shift and d , the vertical shift.

here a

#=1/2,b=1,c=-pi" and "d=0# hence period

#=(2pi)/1=2pi" and amplitude"=|1/2|=1/2#

Amplitude:

The period (pitch) is

#### Explanation:

The given equation starts with the basis of sine. This has a maximum and minimum values of -1 to +1.

Making this into a product of

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I am interpreting this to be the equivalent of pitch. Like that of a thread. That is the distance between maximum values or any other such repeat.

This standard 'untampered with' period is

As we do not have this structure the pitch remains at the standard

Suppose we had

Imagine that we have two graphs. The one we are drawing and a reference one of

As we are considering

We then turn to the graph we are drawing and plot this recorded

So multiplying