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The constant term of the minimal polynomial of $cos\left(2\pi /n\right)$ over ℚ

Fixed Point Theory and Applications20132013:77

https://doi.org/10.1186/1687-1812-2013-77

• Accepted: 11 March 2013
• Published:

Abstract

Let $H\left({\lambda }_{q}\right)$ be the Hecke group associated to ${\lambda }_{q}=2cos\frac{\pi }{q}$ for $q\ge 3$ integer. In this paper, we determine the constant term of the minimal polynomial of ${\lambda }_{q}$ denoted by ${P}_{q}^{\ast }\left(x\right)$.

MSC:12E05, 20H05.

Keywords

• Hecke groups
• minimal polynomial
• constant term

1 Introduction

The Hecke groups $H\left(\lambda \right)$ are defined to be the maximal discrete subgroups of $\mathit{PSL}\left(2,\mathbb{R}\right)$ generated by two linear fractional transformations
$T\left(z\right)=-\frac{1}{z}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}S\left(z\right)=-\frac{1}{z+\lambda },$

where λ is a fixed positive real number.

Hecke  showed that $H\left(\lambda \right)$ is Fuchsian if and only if $\lambda ={\lambda }_{q}=2cos\frac{\pi }{q}$ for $q\ge 3$ is an integer, or $\lambda \ge 2$. In this paper, we only consider the former case and denote the corresponding Hecke groups by $H\left({\lambda }_{q}\right)$. It is well known that $H\left({\lambda }_{q}\right)$ has a presentation as follows (see ):
$H\left({\lambda }_{q}\right)=〈T,S\mid {T}^{2}={S}^{q}=I〉.$
(1)

These groups are isomorphic to the free product of two finite cyclic groups of orders 2 and q.

The first few Hecke groups are $H\left({\lambda }_{3}\right)=\mathrm{\Gamma }=\mathit{PSL}\left(2,\mathbb{Z}\right)$ (the modular group), $H\left({\lambda }_{4}\right)=H\left(\sqrt{2}\right)$, $H\left({\lambda }_{5}\right)=H\left(\frac{1+\sqrt{5}}{2}\right)$, and $H\left({\lambda }_{6}\right)=H\left(\sqrt{3}\right)$. It is clear from the above that $H\left({\lambda }_{q}\right)\subset \mathit{PSL}\left(2,\mathbb{Z}\left[{\lambda }_{q}\right]\right)$, but unlike in the modular group case (the case $q=3$), the inclusion is strict and the index $\left[\mathit{PSL}\left(2,\mathbb{Z}\left[{\lambda }_{q}\right]\right):H\left({\lambda }_{q}\right)\right]$ is infinite as $H\left({\lambda }_{q}\right)$ is discrete, whereas $\mathit{PSL}\left(2,\mathbb{Z}\left[{\lambda }_{q}\right]\right)$ is not for $q\ge 4$.

On the other hand, it is well known that ζ, a primitive n th root of unity, satisfies the equation
${x}^{n}-1=0.$
(2)
In , Cangul studied the minimal polynomials of the real part of ζ, i.e., of $cos\left(2\pi /n\right)$ over the rationals. He used a paper of Watkins and Zeitlin  to produce further results. Also, he made use of two classes of polynomials called Chebycheff and Dickson polynomials. It is known that for $n\in \mathbb{N}\cup \left\{0\right\}$, the n th Chebycheff polynomial, denoted by ${T}_{n}\left(x\right)$, is defined by
${T}_{n}\left(x\right)=cos\left(n\cdot arccosx\right),\phantom{\rule{1em}{0ex}}x\in \mathbb{R},|x|\le 1,$
(3)
or
${T}_{n}\left(cos\theta \right)=cosn\theta ,\phantom{\rule{1em}{0ex}}\theta \in \mathbb{R}\phantom{\rule{0.25em}{0ex}}\left(\theta =arccosx+2k\pi ,k\in \mathbb{Z}\right).$
(4)

Here we use Chebycheff polynomials.

For $n\in \mathbb{N}$, Cangul denoted the minimal polynomial of $cos\left(2\pi /n\right)$ over Q by ${\mathrm{\Psi }}_{n}\left(x\right)$. Then he obtained the following formula for the minimal polynomial ${\mathrm{\Psi }}_{n}\left(x\right)$.

Theorem 1 ([, Theorem 1])

Let $m\in \mathbb{N}$ and $n=\left[|m/2|\right]$. Then
1. (a)

If $m=1$, then ${\mathrm{\Psi }}_{1}\left(x\right)=x-1$, and if $m=2$, then ${\mathrm{\Psi }}_{2}\left(x\right)=x+1$.

2. (b)
If m is an odd prime, then
${\mathrm{\Psi }}_{m}\left(x\right)=\frac{{T}_{n+1}\left(x\right)-{T}_{n}\left(x\right)}{{2}^{n}\left(x-1\right)}.$
(5)

3. (c)
If $4\mid m$, then
${\mathrm{\Psi }}_{m}\left(x\right)=\frac{{T}_{n+1}\left(x\right)-{T}_{n-1}\left(x\right)}{{2}^{n/2}\left({T}_{\frac{n}{2}+1}\left(x\right)-{T}_{\frac{n}{2}-1}\left(x\right)\right){\prod }_{d\mid m,d\ne m,d\mid \frac{m}{2}}^{q-1}{\mathrm{\Psi }}_{d}\left(x\right)}.$
(6)

4. (d)
If m is even and $m/2$ is odd, then
${\mathrm{\Psi }}_{m}\left(x\right)=\frac{{T}_{n+1}\left(x\right)-{T}_{n-1}\left(x\right)}{{2}^{n-{n}^{\mathrm{\prime }}}\left({T}_{{n}^{\mathrm{\prime }}+1}\left(x\right)-{T}_{{n}^{\mathrm{\prime }}}\left(x\right)\right){\prod }_{d\mid m,d\ne m,d\mathit{\text{even}}}^{q-1}{\mathrm{\Psi }}_{d}\left(x\right)},$
(7)

where ${n}^{\mathrm{\prime }}=\frac{\frac{m}{2}-1}{2}$.
1. (e)
Let m be odd and let p be a prime dividing m. If ${p}^{2}\mid m$, then
${\mathrm{\Psi }}_{m}\left(x\right)=\frac{{T}_{n+1}\left(x\right)-{T}_{n}\left(x\right)}{{2}^{n-{n}^{\mathrm{\prime }}}\left({T}_{{n}^{\mathrm{\prime }}+1}\left(x\right)-{T}_{{n}^{\mathrm{\prime }}}\left(x\right)\right)},$
(8)

where ${n}^{\mathrm{\prime }}=\frac{\frac{m}{p}-1}{2}$. If ${p}^{2}\mid m$, then
${\mathrm{\Psi }}_{m}\left(x\right)=\frac{{T}_{n+1}\left(x\right)-{T}_{n}\left(x\right)}{{2}^{n-{n}^{\mathrm{\prime }}}\left({T}_{{n}^{\mathrm{\prime }}+1}\left(x\right)-{T}_{{n}^{\mathrm{\prime }}}\left(x\right)\right){\mathrm{\Psi }}_{p}\left(x\right)},$
(9)

where ${n}^{\mathrm{\prime }}=\frac{\frac{m}{p}-1}{2}$.

For the first four Hecke groups Γ, $H\left(\sqrt{2}\right)$ , $H\left({\lambda }_{5}\right)$, and $H\left(\sqrt{3}\right)$, we can find the minimal polynomial, denoted by ${P}_{q}^{\ast }\left(x\right)$, of ${\lambda }_{q}$ over Q as ${\lambda }_{3}-1$, ${\lambda }_{4}^{2}-2$, ${\lambda }_{5}^{2}-{\lambda }_{5}-1$, and ${\lambda }_{6}^{2}-3$, respectively. However, for $q\ge 7$, the algebraic number ${\lambda }_{q}=2cos\frac{\pi }{q}$ is a root of a minimal polynomial of degree ≥3. Therefore, it is not possible to determine ${\lambda }_{q}$ for $q\ge 7$ as nicely as in the first four cases. Because of this, it is easy to find and study with the minimal polynomial of ${\lambda }_{q}$ instead of ${\lambda }_{q}$ itself. The minimal polynomial of ${\lambda }_{q}$ has been used for many aspects in the literature (see  and ).

Notice that there is a relation
${P}_{q}^{\ast }\left(x\right)={2}^{\phi \left(2q\right)/2}\cdot {\mathrm{\Psi }}_{2q}\left(\frac{x}{2}\right)$

between ${P}_{q}^{\ast }\left(x\right)$ and ${\mathrm{\Psi }}_{m}\left(x\right)$.

In , when the principal congruence subgroups of $H\left({\lambda }_{q}\right)$ for $q\ge 7$ prime were studied, we needed to know whether the minimal polynomial of ${\lambda }_{q}$ is congruent to 0 modulo p for prime p and also the constant term of it modulo p.

In this paper, we determine the constant term of the minimal polynomial ${P}_{q}^{\ast }\left(x\right)$ of ${\lambda }_{q}$. We deal with odd and even q cases separately. Of course, this problem is easier to solve when q is odd.

2 The constant term of ${P}_{q}^{\ast }\left(x\right)$

In this section, we calculate the constant term for all values of q. Let c denote the constant term of the minimal polynomial ${P}_{q}^{\ast }\left(x\right)$ of ${\lambda }_{q}$, i.e.,
$c={P}_{q}^{\ast }\left(0\right).$
(10)
We know from [, Lemma, p.473] that the roots of ${P}_{q}^{\ast }\left(x\right)$ are $2cos\frac{h\pi }{q}$ with $\left(h,q\right)=1$, h odd and $1\le h\le q-1$. Being the constant term, c is equal to the product of all roots of ${P}_{q}^{\ast }\left(x\right)$:
$c=\underset{\underset{h\phantom{\rule{0.25em}{0ex}}\mathrm{odd}}{\underset{\left(h,q\right)=1}{h=1}}}{\prod ^{q-1}}2cos\frac{h\pi }{q}.$
(11)

Therefore we need to calculate the product on the right-hand side of (11). To do this, we need the following result given in .

Lemma 1 ${\prod }_{h=0}^{q-1}2sin\left(\frac{h\pi }{q}+\theta \right)=2sinq\theta$.

We now want to obtain a similar formula for cosine. Replacing θ by $\frac{\pi }{2}-\theta$, we get
$\prod _{h=0}^{q-1}2cos\left(\frac{h\pi }{q}-\theta \right)=2sinq\left(\frac{\pi }{2}-\theta \right).$
(12)
Let now μ denote the Möbius function defined by
(13)
for $n\in \mathbb{N}$. It is known that
(14)
Therefore
$\underset{\underset{\left(h,q\right)=1}{h=0}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}-\theta \right)=\prod _{d\mid q}{\left(sind\left(\frac{\pi }{2}-\theta \right)\right)}^{\mu \left(q/d\right)}.$
(16)
Finally, as $\left(0,q\right)\ne 1$, we can write (16) as
$\underset{\underset{\left(h,q\right)=1}{h=1}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}-\theta \right)=\prod _{d\mid q}{\left(sind\left(\frac{\pi }{2}-\theta \right)\right)}^{\mu \left(q/d\right)}.$
(17)
Note that if q is even, then
$\underset{\underset{\left(h,q\right)=1}{h=1}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}\right)=\underset{\underset{h\phantom{\rule{0.25em}{0ex}}\mathrm{odd}}{\underset{\left(h,q\right)=1}{h=1}}}{\prod ^{q-1}}2cos\frac{h\pi }{q}=c,$
(18)
while if q is odd, then
$|\underset{\underset{\left(h,q\right)=1}{h=1}}{\prod ^{q-1}}2cos\left(\frac{h\pi }{q}\right)|={c}^{2},$
(19)
as $cos\left(h-i\right)\frac{\pi }{q}=-cos\frac{i\pi }{q}$. Also note that
(20)

To compute c, we let $\theta \to 0$ in (17). If d is odd, then $sind\left(\frac{\pi }{2}-\theta \right)\to ±1$ as $\theta \to 0$ by (20). So, we are only concerned with even d. Indeed, if q is odd, then the left-hand side at $\theta =0$ is equal to ±1. Therefore we have the following result.

Theorem 2 Let q be odd. Then
$|c|=1.$
(21)

Proof It follows from (19) and (20). □

Let us now investigate the case of even q. As $\left(h,q\right)=1$, h must be odd. So, by a similar discussion, we get the following.

Theorem 3 Let q be even. Then
$c=\underset{\theta \to 0}{lim}\prod _{d\mid q}{\left(sind\left(\frac{\pi }{2}-\theta \right)\right)}^{\mu \left(q/d\right)}.$
(22)
Proof Note that by (20), the right-hand side of (22) becomes a product of $±{\left(cosd\theta \right)}^{±1}$s and $±{\left(sind\theta \right)}^{±1}$s. Above we saw that we can omit the former ones as they tend to ±1 as θ tends to 0. Now, as ${\sum }_{d\mid n}\mu \left(d\right)=0$, there are equal numbers of the latter kind factors in the numerator and denominator, i.e., if there is a factor $sind\theta$ in the numerator, then there is a factor $sin{d}^{\mathrm{\prime }}\theta$ in the denominator. Then using the fact that
$\underset{\theta \to 0}{lim}\frac{sink\theta }{sinl\theta }=\frac{k}{l},$
(23)

we can calculate c.

In fact the calculations show that there are three possibilities:

(i) Let $q={2}^{{\alpha }_{0}}$, ${\alpha }_{0}\ge 2$. Then the only divisors of q such that $\mu \left(q/d\right)\ne 0$ are $d={2}^{{\alpha }_{0}}$ and ${2}^{{\alpha }_{0}-1}$. Therefore
(24)
(ii) Secondly, let $q=2{p}^{\alpha }$, $\alpha \ge 1$, p odd prime. Then the only divisors of q such that $\mu \left(q/d\right)\ne 0$ are $d=2{p}^{\alpha }$, $2{p}^{\alpha -1}$, ${p}^{\alpha }$ and ${p}^{\alpha -1}$. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin2{p}^{\alpha }\left(\frac{\pi }{2}-\theta \right)\cdot sin{p}^{\alpha -1}\left(\frac{\pi }{2}-\theta \right)}{sin{p}^{\alpha }\left(\frac{\pi }{2}-\theta \right)\cdot sin2{p}^{\alpha -1}\left(\frac{\pi }{2}-\theta \right)}\\ =& \underset{\theta \to 0}{lim}ϵ\cdot \frac{sin2{p}^{\alpha }\theta \cdot cos{p}^{\alpha -1}\theta }{cos{p}^{\alpha }\theta \cdot sin2{p}^{\alpha -1}\theta }\\ =& ϵ\cdot p,\end{array}$
(25)
where
(26)
(iii) Let q be different from above. Then q can be written as
$q={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}},$
(27)

where ${p}_{i}$ are distinct odd primes and ${\alpha }_{i}\ge 1$, $0\le i\le k$.

Here we consider the first two cases $k=1$ and $k=2$.

Let $k=1$, i.e., let $q={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}$. We have already discussed the case ${\alpha }_{0}=1$. Let ${\alpha }_{0}>1$. Then the only divisors d of q with $\mu \left(q/d\right)\ne 0$ are $d={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}$, ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}$, ${2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}$ and ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}$. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}\left(\frac{\pi }{2}-\theta \right)}{sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}\left(\frac{\pi }{2}-\theta \right)}\\ =& 1.\end{array}$
(28)
Now let $k=2$, i.e., let $q={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}$, (${p}_{1}<{p}_{2}$). Similarly, all divisors d of q such that $\mu \left(q/d\right)\ne 0$ are $d={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}$, ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}$, ${2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}$, ${2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}-1}$, ${2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}$, ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}$, ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}-1}$ and ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}$. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}-1}\left(\frac{\pi }{2}-\theta \right)}{sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}-1}\left(\frac{\pi }{2}-\theta \right)}\\ ×\underset{\theta \to 0}{lim}\frac{sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\left(\frac{\pi }{2}-\theta \right)}{sin{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}\left(\frac{\pi }{2}-\theta \right)\cdot sin{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\left(\frac{\pi }{2}-\theta \right)}\\ =& 1.\end{array}$
(29)
Finally, $k\ge 3$, i.e., let
In this case the proof is similar, but rather more complicated. In fact, the number of all divisors d of q such that $\mu \left(q/d\right)\ne 0$ is ${2}^{k+1}$. There is $\left(\begin{array}{c}k+1\\ 0\end{array}\right)=1$ divisor of the form
$d={2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}}.$
There are $\left(\begin{array}{c}k+1\\ 1\end{array}\right)=k+1$ divisors of the form
$d={2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}},{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}\cdots {p}_{k}^{{\alpha }_{k}},\dots ,{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}-1}.$
There are $\left(\begin{array}{c}k+1\\ 2\end{array}\right)=\frac{k\left(k+1\right)}{2}$ divisors of the form
$\begin{array}{rcl}d& =& {2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{k}^{{\alpha }_{k}},{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}-1}\cdots {p}_{k}^{{\alpha }_{k}},\dots ,{2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{k}^{{\alpha }_{k}-1},\\ {2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}\cdots {p}_{k}^{{\alpha }_{k}},\dots ,{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}}\cdots {p}_{k}^{{\alpha }_{k}-1},\dots ,{2}^{{\alpha }_{0}}{p}_{1}^{{\alpha }_{1}}\cdots {p}_{k-1}^{{\alpha }_{k-1}-1}{p}_{k}^{{\alpha }_{k}-1}.\end{array}$

If we continue, we can find other divisors d of q, similarly. Finally, there is $\left(\begin{array}{c}k+1\\ k+1\end{array}\right)=1$ divisor of the form ${2}^{{\alpha }_{0}-1}{p}_{1}^{{\alpha }_{1}-1}{p}_{2}^{{\alpha }_{2}-1}\cdots {p}_{k}^{{\alpha }_{k}-1}$. Thus, the product of all coefficients d in the factors $sind\left(\frac{\pi }{2}-\theta \right)$ in the numerator is equal to the product of all coefficients e in the factors $sine\left(\frac{\pi }{2}-\theta \right)$ in the denominator implying $c=1$. Therefore the proof is completed. □

Now we give an example for all possible even q cases.

Example 1 (i) Let $q=8={2}^{3}$. The only divisors of 8 such that $\mu \left(8/d\right)\ne 0$ are $d=8$ and 4. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin8\left(\frac{\pi }{2}-\theta \right)}{sin4\left(\frac{\pi }{2}-\theta \right)}\\ =& 2.\end{array}$
(ii) Let $q=14=2\cdot 7$. The only divisors of 14 such that $\mu \left(14/d\right)\ne 0$ are $d=14,2,7$ and 1. Therefore
$\begin{array}{rcl}c& =& ϵ\cdot \underset{\theta \to 0}{lim}\frac{sin14\left(\frac{\pi }{2}-\theta \right)\cdot sin\left(\frac{\pi }{2}-\theta \right)}{sin7\left(\frac{\pi }{2}-\theta \right)\cdot sin2\left(\frac{\pi }{2}-\theta \right)}\\ =& -7,\end{array}$

since $p\equiv -1mod4$.

(iii) Let $q=24={2}^{3}\cdot 3$. The only divisors of 24 such that $\mu \left(24/d\right)\ne 0$ are $d=24,12,8$ and 4. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin24\left(\frac{\pi }{2}-\theta \right)\cdot sin4\left(\frac{\pi }{2}-\theta \right)}{sin12\left(\frac{\pi }{2}-\theta \right)\cdot sin8\left(\frac{\pi }{2}-\theta \right)}\\ =& 1.\end{array}$
(iv) Let $q=30=2\cdot 3\cdot 5$. The only divisors of 30 such that $\mu \left(30/d\right)\ne 0$ are $d=30,15,10,6,5,3,2$ and 1. Therefore
$\begin{array}{rcl}c& =& \underset{\theta \to 0}{lim}\frac{sin\left(\frac{\pi }{2}-\theta \right)\cdot sin6\left(\frac{\pi }{2}-\theta \right)\cdot sin10\left(\frac{\pi }{2}-\theta \right)\cdot sin15\left(\frac{\pi }{2}-\theta \right)}{sin2\left(\frac{\pi }{2}-\theta \right)\cdot sin3\left(\frac{\pi }{2}-\theta \right)\cdot sin5\left(\frac{\pi }{2}-\theta \right)\cdot sin30\left(\frac{\pi }{2}-\theta \right)}\\ =& 1.\end{array}$

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Both authors are supported by the Scientific Research Fund of Uludag University under the project number F2012/15 and the second author is supported under F2012/19.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Arts and Science, Uludag University, Gorukle Campus, Bursa, 16059, Turkey

References 