How do you graph #y=1/4sin(2x+2pi)#?

1 Answer
Dec 9, 2017

Calculate the period from #b# (the #2# in #2x#), divide the period into 4 angles, solve the equation for the 4 angles, plot the solutions, then estimate the rest of the graph.

Explanation:

The general form is #f(x) = a sin(bx - c) + d#. List down what is known (when it comes to sines and cosines, a full cycle is #2π#.):

#a# (amplitude) #= 1/4#

#b# ("cycles per full cycle") #= 2#

Period (radians in a cycle) = #("full cycle")/b = (2π)/2 = π# radians

#c# (phase shift) #= -2π# radians

#d# (vertical shift) #= 0#

What we'll need here is the period. Since the period is #π# radians, we obtain several angles to solve by dividing that into four (plus #0#):
#0, π/4, π/2, (3π)/4, π#

The reason we divided by four is so that we can reach all "four corners": the midline, the highest point, the midline again, and then the lowest point. Values in between them could later be estimated.

Shifting by a full cycle, regardless of the direction, does not affect the graph, so the phase shift could also just be #0# radians and that it does nothing. Vertical shift is supposed to shift the "midline" of the graph vertically, but here it is #0#.

Now input each angle into the function and obtain the results:

#f(0) = 1/4 sin(2(0)) = 1/4 sin(0) = 1/4 (0) = 0#

#f(π/4) = 1/4 sin(2(π/4)) = 1/4 sin(π/2) = 1/4 (1) = 1/4#

#f(π/2) = 1/4 sin(2(π/2)) = 1/4 sin(π) = 1/4 (0) = 0#

#f((3π)/4) = 1/4 sin(2(3π/4)) = 1/4 sin(3π/2) = 1/4 (-1) = -1/4#

#f(π) = 1/4 sin(2(π)) = 1/4 sin(2π) = 1/4 (0) = 0#

Now that we have these, let's plot them! It should look something like this:

graph{((x - 0)^2 + (y - 0)^2 - (0.05)^2)((x - pi/4)^2 + (y - 1/4)^2 - (0.05)^2)((x - pi/2)^2 + (y - 0)^2 - 0.05^2)((x - (3pi)/4)^2 + (y + 1/4)^2 - 0.05^2)((x - pi)^2 + (y - 0)^2 - 0.05^2) = 0 [-1.421, 3.882, -1.302, 1.35]}

Now estimate the graph:

graph{((x - 0)^2 + (y - 0)^2 - 0.05^2)((x - pi/4)^2 + (y - 1/4)^2 - 0.05^2)((x - pi/2)^2 + (y - 0)^2 - 0.05^2)((x - 3pi/4)^2 + (y + 1/4)^2 - 0.05^2)((x - pi)^2 + (y - 0)^2 - 0.05^2)(y - 1/4 sin(2x)) = 0 [-0.305, 4.764, -1.144, 1.39]}

And continue the cycle, both forwards and backwards!

graph{1/4 sin(2x)}

There you have it!