Let $q$ be an element of $\mathbb{Q}$ (rational numbers). How can I prove that $\sin (2\pi q)$ is algebraic over $\mathbb{Q}$ for any $q$? I am trying the method: Euler formula: $e^{i\theta} =\cos \theta+i\sin \theta$ so $2\sin\theta = ie^{i\theta}-ie^{-i\theta}$. $i$ is root of $x^2+1$. So if I prove that $e^{i\theta}$ and $e^{-i\theta}$ is algebraic over $\mathbb{Q}$f or $\theta=2\pi q$, it will be ok.
122900
Here $p$ and $q$ are co-prime integers. Is it necessarily an algebraic number?
99574
How do I go about proving this? I know one method is:  $\eqalign{    \cos (90^\circ + \theta ) &amp;\equiv \cos90^\circ \cos\theta  - \sin90^\circ \sin\theta   \cr    &amp;  \equiv (0)(\cos\theta ) - (1)(\sin \theta )  \cr    &amp;  \equiv  - \sin \theta  \cr} $  I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.  Thanks alot.
91573
I want to prove this without using any of the properties about the field of algebraic numbers (specifically that it is one). Essentially I just want to find a polynomial for which $\cos\frac{2\pi}{n}$ is a root.  I know roots of unity and De Moivre's theorem is clearly going to be important here but I just can't see how to actually construct the polynomial from these facts.
102648
I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible.
66462
I am trying to solve the following two problems:  1) Prove that the functions $\cos(z)$, $\sin(z)$ are surjective over the complex numbers.  2) Find all $z \in \mathbb C$: $cos(z) \in \mathbb R$ and find all $z \in \mathbb C$: $\sin(z) \in \mathbb R$.  For 1),I've tried to prove it for the function $\cos(z)$ (I suppose the other one is analogue)  so I've used the fact that $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$.  Let $w \in \mathbb C$, I want to show there exists $z \in \mathbb C : f(z)=w$, i.e., $\dfrac{e^{iz}+e^{-iz}}{2}=w$, multiplying by $2$ and then by $e^{iz}$ yields $e^{2iz}+1=2we^{iz}$ iff $e^{2iz}-2we^{iz}+1=0$.   I don't know if this approach is the correct one but here I've replaced $e^{iz}$ by $x$, so the solutions of the equation would be the roots of the polynomial $p(x)=x^2-2wx+1$, by the quadratic formula, I get that $x \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$, where $w_0^2=4w^2-4$.  Then, $e^{iz} \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$. At this point I got lost, I would like to explicitly show that $z$ exists and I can't see existence directly from the fact that $e^{iz}=\dfrac{2w+w_0}{2}$ or $e^{iz}=\dfrac{2w-w_0}{2}$.  First I thought of taking logarithm of both sides of the equation ir order to solve for $z$, but this is not a legitimate operation unless $e^{iz} \in \mathbb R$.   I couldn't go any farther.  For point 2) I have no idea what to do, should I use the identity $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$?  I would appreciate some help with the two points (specially with point 2), at least in 1) I could do something).   Btw, Happy new year!
46357
I gave an exercise to a student the other day, thinking I had a simple solution for it, but it seems that my solution was just bullshit (at least, not conclusive). Well, at least, I warned her I wasn't sure of my solution :-)  Here it is : $a\in[-1,1]$, show that all roots of polynomial $P=X^{n+1}-aX^n+aX-1$ have modulus $1$ (I try different ways of expressing the problem, tell me which one is better english...).  It is easy to see that    roots of $P$ are $1$, $-1$ for some special cases, and complex not real roots $z_i$ such that $\overline{z_i}$ is also a root, as $P$ is kind of reciprocal, $z$ root implies $\frac1z$ is also a root$.   All the other things I tried are not conclusive. For example, $a\in[-1,1]$ may be translated as $a=\cos\theta$ for some $\theta\in\mathbb R$, but I don't see how to use this. I tried some rewriting, but nothing seems to work. I tried to work on the modulus of a root, tried the relations roots-coefficients...  I'm quite stuck here. Could you help me please ?  Thanks.
95597
So obviously using quadractic formula, we have three cases for the roots which depends on the discriminant. I am not sure if I am right on this but,  Case 1: $a^{2}-4b&gt;0$.   Then we have two distinct real roots so this factors as $$\mathbb{R}[x]/(x-\alpha) \oplus \mathbb{R}[x]/(x-\beta).$$ (My guess here is since $\alpha$ and $\beta$ are different, these two ideals are comaximal so I can apply the Chinese Remainder Theorem). But this is isomorphic to $$\mathbb{R} \oplus \mathbb{R}.$$  Case 2: $a^{2}-4b &lt; 0$.  In this case, we have a complex root so we have an extension of degree $2$. But obviously whatever root we have, this is contained in $\mathbb{R}[i]\cong \mathbb{C}$ so we must have this is $\mathbb{C}$ by the tower law since $\mathbb{R}[i]$ is also a degree $2$ extension of $\mathbb{R}$.  Case 3: $a^{2}-4b=0$.  Then we have a real root with multiplicity $2$. I am not sure what this is as if we have the root is $\alpha$, we get $$\mathbb{R}[x]/(x-\alpha)^{2}$$ Since these roots are the same, $(x-\alpha)$ and $(x-\alpha)$ are not comaximal I believe so I don't know what this ring is.
72513
$$\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(4x)\cdots\sin(2^kx)\over x^{k+1}}\mathrm dx=2^{0.5(k^2-k-2)}\pi\tag1$$  $k\ge0$  Experimental using wolfram integrator, let me to conclude the closed form for $(1)$ is as follow above.  But I don't know how shows it.  How can we prove $(1)$?  We have  $$\prod_{j=0}^{k-1}\cos(2^jx)={\sin(2^kx)\over 2^k\sin(x)}\tag2$$  I am sure how is this product will relate to $(1)$  Setting $k=0$ we have the   $$\int_{0}^{\infty}{\sin(x)\over x}\mathrm dx={\pi\over 2}$$
59687
Prove $\sin(\pi/2)=1$ using the Taylor series definition of $\sin x$,   $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$   It seems rather messy to substitute in $\pi/2$ for $x$. So we have  $$\sin(\pi/2)=\sum_{n=0}^{\infty} \frac{(-1)^n(\pi/2)^{2n+1}}{(2n+1)!}.$$  I'm not too sure where to go from here.  Any help would be appreciated! Thanks!
77990
The proof I'm familiar with that the algebraic numbers $\mathbb A$ form a field uses the fact that the resultant of two polynomials $p,q\in\mathbb Q[x]$ satisfies the following properties:   It is $0$ iff $p$ and $q$ have a common factor. It is a polynomial in the coefficients of $p$ and $q$.   We then introduce a new variable and cleverly manipulate $p$ and $q$ to get polynomials which vanish at the sums and products of their roots. This is in some ways a nice proof, e.g. it is constructive and so can be converted into an algorithm to find such polynomials (which I in fact just finished doing in C). But I don't find it very enlightening; it seems like the fact that $\mathbb A$ is a field is simply an accident. Is there a more enlightening proof of this fact?
8836
Eulers product identity is as follows  \begin{align} \prod_{n=1}^{\infty} (1-q^{2n-1}) (1+q^{n}) =1 \end{align}  How one can explicitly prove this identity?  Note here $q$ deonotes a complex number satisfying $|q|&lt;1$
82453
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.   I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
90759
Prove that $$\cos(\pi/11)+\cos(3\pi/11)+\cos(5\pi/11)+\cos(7\pi/11)+\cos(9\pi/11)=1/2$$ using Euler's formula.  Everything I tried has failed so far.  Here is one thing I tried, but obviously didn't work. $$\Re e \{e^{\frac{\pi}{11}i}(1+e^{\frac{2\pi}{11}i}+e^{\frac{4\pi}{11}i}+e^{\frac{6\pi}{11}i}+e^{\frac{8\pi}{11}i}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{e^{2\pi i}}+\sqrt[11]{e^{4\pi i}}+\sqrt[11]{e^{6\pi i}}+\sqrt[11]{e^{8\pi i}}) \}=\frac{1}{2}$$ $$\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}) \}=\frac{1}{2}$$ $$\Re e \{5e^{\frac{\pi}{11}i} \}=\frac{1}{2}$$ $$5\cos(\frac{\pi}{11})=\frac{1}{2}$$ Which isn't true :D  Thanks in advance
79764
I would to prove that  $$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+  \frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$ given that  $$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$ where $z\neq 1$.  I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
121695
Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$  as a root).  I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to  square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals.
21199
I want to prove that     $p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.   My attempt.   GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma). Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.   Does this proof is correct?  UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?
66507
How does one go about proving that the sums and products of two algebraic numbers over a field $F$ (say $a,b\in K$, where $K/F$ is a field extension) is also algebraic?   If we call $f_a$ and $f_b$ the min. poly's of $a$ and $b$, then I'm assuming the answer involves such polynomials. Perhaps looking at their roots in splitting fields for both of them? And finding a "big" splitting field, constructed from those two other ones?      In particular, I'd like a way of explicitly constructing the minimal polynomials $\ f_{ab}$ of $ab$ and $f_{a+b}$ of $a+b$.     I read somewhere that $g(x)=\Pi_j\Pi_i (x-\alpha_i\beta_j)$ works for $ab$, where the $\alpha_i$ and $\beta_j$ are the roots of $f_a$ and $f_b$, respectively, but I do not know why $g(x)\in F[x]$.  Similar remarks for $a+b$