What is the graph of #y=sin(x/2)#?

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Answer 1 · 2014-10-06T05:00:09

First, calculate the period.

#omega=(2pi)/B=(2pi)/(1/2)=((2pi)/1)*(2/1)=4pi#

Break up #6pi# into fourth by dividing by #4#.

#(4pi)/(4)=pi#

#0,pi,2pi,3pi,4pi ->x#-values

These #x# values correspond to ...

#sin(0)=0#

#sin((pi)/(2))=1#

#sin(pi)=0#

#sin((3pi)/2)=-1#

#sin(2pi)=0#

Enter the function using the Y= button

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Press the WINDOW button.
Enter the Xmin of #0# and Xmax of #4pi#.
The calculator converts #4pi# to its decimal equivalent.

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Press the GRAPH button.

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