How do you find the amplitude and period of #y=1/2sin2theta#?

1 Answer
Jan 25, 2017

#"amplitude " =1/2," period" =pi#

Explanation:

The standard form of the #color(blue)"sine function"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=asin(bx+c)+d)color(white)(2/2)|)))#

#"where amplitude "=|a|," period "=(2pi)/b#

#"phase shift "=-c/b" and vertical shift " =d#

#"here "a=1/2,b=2,c=0,d=0#

#rArr"amplitude "=|1/2|=1/2," period " =(2pi)/2=pi#