Find the distance between (5 , 1 ) and ( 3 ,6)
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The half-range sine series is the expansion for 
with the assumption that is considered to be an odd function over its full range, 
. So for (a), you're essentially finding the full range expansion of the function
with period 2 so that
for 
and integers 
.
Now, since is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with
where
In this case,
, so
The half-range sine series expansion for is then
which can be further simplified by considering the even/odd cases of, but there's no need for that here.
The half-range cosine series is computed similarly, this time assuming is even/symmetric across its full range. In other words, you are finding the full range series expansion for
Now the sine series expansion vanishes, leaving you with
where
for
. Again, . You should find that
Here, splitting into even/odd cases actually reduces this further. Notice that when is even, the expression above simplifies to
while for odd, you have
So the half-range cosine series expansion would be
Attached are plots of the first few terms of each series overlaid onto plots of. In the half-range sine series (right), I use 
terms, and in the half-range cosine series (left), I use 
or 
terms. (It's a bit more difficult to distinguish from the latter because the cosine series converges so much faster.)
with period 2 so that
Now, since
where
In this case,
The half-range sine series expansion for
which can be further simplified by considering the even/odd cases of
The half-range cosine series is computed similarly, this time assuming
Now the sine series expansion vanishes, leaving you with
where
for
Here, splitting into even/odd cases actually reduces this further. Notice that when
while for odd
So the half-range cosine series expansion would be
Attached are plots of the first few terms of each series overlaid onto plots of