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I was just wondering if there is any way to get an exact expression (with radicals) for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$. In case it's relevant, I want to express $z = \sqrt[4]{8} e^{\frac{5\pi}{8}i}$ in binomial form, and I know that

$$ \begin{aligned} \cos \frac{5\pi}{8} &= - \cos \frac{3\pi}{8} \\ \sin \frac{5\pi}{8} &= \sin \frac{3\pi}{8} \end{aligned} $$

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What's not exact about $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$? (Okay, okay, use the half-angle formulas.) – Qiaochu Yuan Mar 13 '12 at 2:39
These wolframalpha answers are lame. It's like some kind of cargo cult. Who cares what WA says the answer is if we have no way of seeing how the answer was derived and no way of knowing whether it's correct? Posting this kind of thing takes zero effort and zero understanding, and it gives zero benefit to anyone. – Ben Crowell Mar 13 '12 at 3:01
@Ben: I don't think that's true. It can be quite helpful psychologically to know that a question has a nice answer, and often that's enough to lead someone to an independent answer to the question. – Qiaochu Yuan Mar 13 '12 at 3:12
In any case, @J.D., you don't need WolframAlpha to tell you that. $\cos r \pi$ is always algebraic for rational $r$ since it equals $\frac{e^{i r \pi} + e^{-i r pi}}{2}$, the number $e^{i r \pi}$ satisfies the polynomial $x^q = e^{i p \pi} = (-1)^p$ where $r = \frac{p}{q}$, and the sum of two algebraic numbers is algebraic (and similarly for sines). One can explicitly write down a polynomial satisfied by cosines (and sines) using the Chebyshev polynomials ( – Qiaochu Yuan Mar 13 '12 at 3:27
@BenCrowell that's why we leave it as a comment, not as an answer. This is a collaborative website. All relevant information are relevant, and it will benefit future visitors of this post to see every possible relevant solutions and answers in one place, including W|A output. – user2468 Mar 13 '12 at 3:36
up vote 4 down vote accepted

Yes, we can use the double angle formula

$$ 2\cos^2 x - 1 = \cos 2x$$

Pick $x = \frac{3\pi}{8}$ and you can solve it.

Once you have the cos value, you can easily get the sin value.

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Well, that was simple :). Thanks. – Javier Mar 13 '12 at 3:02
@JavierBadia: You are welcome! – Aryabhata Mar 13 '12 at 3:12

The following is a beautiful old theorem of Gauss. Let $n$ be a positive integer. Then there is an expression for $\cos(2\pi/n)$ (and thus for $\sin(2\pi/n)$) in terms of the ordinary arithmetical operations augmented by $\sqrt{\hphantom{aa}}$ (of non-negative quantities) if and only if $n$ is of the shape $$n=2^e p_1p_2\cdots p_k,$$ where $e$ is a non-negative integer, and $\{p_1,p_2,\dots,p_k\}$ are distinct Fermat primes. (There might be no Fermat primes in the collection.)

The Fermat primes are the primes of the shape $2^{2^m}+1$. There are only five Fermat primes known at this time. They are $3$, $5$, $17$, $257$, and $65537$.

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