Cubic and Quartic Formulae

[Manolo Sobrino]

Did you really try that or it's just guessing? It's a dead end. Putting aside the fact that the whole endeavour is a bit futile, the root of the equation that gives you the value of sin(1°) (I don't understand what makes Pi/180 so appealing) has cube roots of a complex number with parts made up of square roots of sums of square roots of some integers. Unfortunately, the "full Reals" Cardan root is not the good one. How do you get a standard pretty closed form for that without the closed forms of values of trigonometric functions with very ugly arguments... that you don't have?

Well, you could just write it down and so be it. After some cranking, a form happens to be:

\begin{equation}
sin\left(1°\right)=\frac{2 \sqrt[3]{2} \left(-1+i \sqrt{3}\right)-\left(1+i \sqrt{3}\right)
\left(-\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2 \sqrt{5}}}+i
\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2 \sqrt{5}}}\right)^{2/3}}{4\ 2^{2/3}
\sqrt[3]{-\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2 \sqrt{5}}}+i
\sqrt{8+\sqrt{3}+\sqrt{15}+\sqrt{10-2 \sqrt{5}}}}}
\end{equation}

I can't resist the version with prettier numbers:

\begin{equation}
sin\left(1°\right)=\frac{-2+2 i \sqrt{3}-\left(1+i \sqrt{3}\right) \left(-\sqrt{4-\sqrt{7+\sqrt{5}+\sqrt{6
\left(5+\sqrt{5}\right)}}}+i \sqrt{4+\sqrt{7+\sqrt{5}+\sqrt{6
\left(5+\sqrt{5}\right)}}}\right)^{2/3}}{4 \sqrt{2}
\sqrt[3]{-\sqrt{4-\sqrt{7+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}}}+i
\sqrt{4+\sqrt{7+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}}}}}
\end{equation}

(Mathematica principal roots.) Yes, the imaginary part is zero.

It is a closed form, but can we compute with just easy radicals of real numbers? No. What did we improve? Nothing.

Of course, with such a closed expression for sin(1°) there are equally computationally inefficient straightforward ways to get such closed expressions for every degree. Those things might look lovely, but what's the point? (No Tim, selling calculators is not a valid point . I'm staying out of the Nspire thing so I don't really get the gist of your question.)


When Bernard talks about Algebra (and Maths in general) I always take notes. This is a great forum, not only about calculators. Sometimes the mood is a bit too upfront if you're not used to it, but you know these issues are so highly controversial...


(edit - fixed some behaviour of the Mathematica TeX parser... Using the roots of i, there are shorter ones indeed:

\begin{equation}
sin\left(1°\right)=\frac{1}{4} (-1)^{59/180} \left(i-2 (-1)^{8/45}+\sqrt{3}\right)
\end{equation}

Eventually you'd get the trivial WAlpha expressions that come straight from the definition of sin (1°) in terms of complex exponentials.

\begin{equation}
sin\left(1°\right)=-\frac{1}{2} (-1)^{89/180} \left(\sqrt[90]{-1}-1\right)
\end{equation}