What is the amplitude of #f(x)=4sin(x)cos(x)#?

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Answer 1 · 2015-02-03T09:53:05

The answer is: #2#.

The amplitude of a periodic function is the numer that multiply the function itself.
Using the double-angle formula of sinus, that says:

#sin2alpha=2sinalphacosalpha#,

we have:

#y=2*2sinxcosx=2sin2x#.

So the amplitude is #2#.

This is the sinus function:

graph{sinx [-10, 10, -5, 5]}

This is the #y=sin2x# function (the period becomes #pi#):

graph{sin(2x) [-10, 10, -5, 5]}

and this is the #y=2sin2x# function:

graph{2sin(2x) [-10, 10, -5, 5]}