Cosine , sine , tangent of rational multiples of π
by adityaguharoy, Feb 1, 2017, 5:07 AM
Lemma:
If the polynomial equation
has a non zero rational root
with
then
and
.
Proof of the Lemma
Corollary :
If a polynomial equation
has a non zero rational root
with
and with
then,
and
.
Proving this corollary immediately
Theorem :
Let
be a rational mutiple of
. Then
,
,
are irrational numbers except when
.
Proof
Problem Solving with the above theorem
See here :
https://www.artofproblemsolving.com/community/c140h563991_2013_pumac_team_16
Is
rational ?
Answer
Solution
A few words
If the polynomial equation





Proof of the Lemma
Replacing
by
in
and then multiplying by
we see that
must be an integer and then using
we get
.
Similarly repalcing
by
in
and then multiplying with
we see that
must be an integer and then using
we get
.
This completes the proof of the lemma.







Similarly repalcing







This completes the proof of the lemma.
Corollary :
If a polynomial equation






Proving this corollary immediately
This immediately follows from the result of the lemma.
Theorem :
Let






Proof
Let
be any positive integer. First, we prove by induction that there is a polynomial
of degree
with integer coefficients and leading coefficient
such that
for all real numbers
.
We note that
and because of :
we have
.
Now, the identity
is easily established using elementary trigonometry.
Thus, we effectively get ,
.
This proves our claim by induction.
Next, let the positive integer
be chosen such that
is also an integer.
With
it follows that :
where the plus sign holds when
is even and the minus sign holds when
is odd.
Thus,
is a solution of
.
Setting aside the cases when
and then applying the above Corollary we conclude that
if rational must be a nonzero integer.
Further we know that ,
and thus
and thus we get the desired result for
function.
For
we use the identity
to get the desired result for
function.
For
we use the identity
and then consider the cases when
is rational and conclude the result for
function because whenver
is rational
must also be rational and we already know the result for
function (the cosine function).
All these work prove the theorem effectively.






We note that



Now, the identity

Thus, we effectively get ,

This proves our claim by induction.
Next, let the positive integer


With




Thus,


Setting aside the cases when


Further we know that ,



For



For







All these work prove the theorem effectively.
Problem Solving with the above theorem
See here :
https://www.artofproblemsolving.com/community/c140h563991_2013_pumac_team_16
Is

Answer
No
Solution
Since,
is a rational multiple of
so we must have
is rational if and only if it is equal to
or
or
or
or
.
But we clearly have these are not possible.
This completes the solution.








But we clearly have these are not possible.
This completes the solution.
A few words
This was the first Princeton Competition problem I could solve.
Discuss when all can
be rational ?
Discuss when all can

This post has been edited 5 times. Last edited by adityaguharoy, Feb 13, 2017, 7:07 AM