How do you find the amplitude and period of #y=1/2cos(3/4theta)#?

1 Answer
Jim G.
Feb 10, 2017

#"amplitude "=1/2," period "=(8pi)/3#

Explanation:

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=acos(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=3/4,c=0,d=0#

#rArr"amplitude "=|1/2|=1/2" and period "=(2pi)/(3/4)=(8pi)/3#

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