How do you show by substitution that the point (π /2, 1) lies on the curves of : #y=2sin(x/3)#?

1 Answer
Alan P.
Oct 6, 2015

Show that #(x,y)=(pi/2,1)# is a solution for #y=2sin(x/3)# by substituting #pi/2# for #x# and solving for #y#

Explanation:

If
#color(white)("XXX")y=2sin(x/3)#
when
#color(white)("XXX")x=pi/2#
then
#color(white)("XXX")y=2sin((pi/2)/3)#

#color(white)("XXX")=2sin(pi/6)#

#color(white)("XXX")=2*1/2#

#color(white)("XXX")=1#

Therefore #(pi/2,1)# is a solution for #y=2sin(x/3)#

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