How do you graph #y=1/2(1-cosx)#?

1 Answer
KillerBunny
Sep 24, 2015

Here's the graph:

graph{1/2(1-cos(x)) [-10, 10, -5, 5]}

Explanation:

You only need to understand which changes were made, starting from the function #cos(x)# (of which I'll assume you know the behavior, and thus the graph), and then to understand what these changes mean. The steps are the following:

  1. Change sign: #cos(x) -> -cos(x)#
  2. Add #1#: #-cos(x) -> 1-cos(x)#
  3. Divide everything by #2#: #1-cos(x) -> 1/2(1-cos(x))#.

Changing the sign of a function simply means to reflect it, with respect to the #x#-axis. So, the change from #cos(x)# to #-cos(x)# is the following:

#cos(x)#:
graph{cos(x) [-12.66, 12.65, -6.33, 6.33]}

#-cos(x)#:
graph{-cos(x) [-12.66, 12.65, -6.33, 6.33]}

Adding a positive constant means to translate the graph upwards. In your case, you'll translate the graph of #-cos(x)# one unit above, obtaining the following:

#1-cos(x)#:
graph{1-cos(x) [-12.66, 12.65, -6.33, 6.33]}

Finally, dividing by #2# "compresses" the function vertically, obtaining the final result.

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