How do you sketch one cycle of #y=1/2cos(1/3x)#?

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Answer 1 · 2018-07-28T04:46:11

See graph and details.

Cycle period for

#y = 1/2 cos (x/3 ) in [ - 1/2, 1/2 ] # is #(2pi)/(1/3) = 6pi#.

From any x = a, one cycle is # ( a + 6pi )#.

Grapm for the cycle# x in [ - 3 pi, 3 pi ]#, with all aspects:
graph{(y-1/2cos(x/3))(y^2-0.25)(x^2-9(pi)^2)=0[-10 10 -2 2]}

See the effect of the scale factor 1/2 on wave amplitude, from the

graph of #y = cos ( x/3 )#.

graph{(y-cos(x/3))=0[-10 10 -2 2]}