Suppose I was working in radians and took

$$\sin \left(\frac{a\pi}{b}\right)$$

where both $a$ and $b$ are integers.

Is there a proof that the output of this function cannot be transcendental? Or, conversely, that

$$\arcsin \left(t \right)$$

where $t$ is some transcendental number, e.g., $e$, cannot be a rational angle $\left(\frac{a\pi }{b}\right)$?

Sadly, I couldn’t get very far with this problem with my maths teacher so any information on the topic would be very useful.

**Answer**

To start with, let’s notice that we might just as well prove this fact for the cosines of rational multiples of $\pi$ as the sines. We just need to use the identity $\sin x=\cos\left(\frac{\pi}{2} – x\right)$, which gives us:

$$\sin \frac{a\pi}{b}=\cos\left(\frac{\pi}{2} – \frac{a\pi}{b}\right)=\cos\left(\frac{(b-2a)\pi}{2b}\right)$$

As this is the cosine of a rational multiple of $\pi$, we get the result for sines more or less for free from the result for cosines.

Once this preliminary is out of the way, the key thing to notice is that $\cos(nx)$ can always be written as a polynomial function of $\cos x$. For example:

\begin{align}

\cos 2x &= 2 \cos^2 x – 1 \\

\cos 3x &= 4 \cos^3 x – 3 \cos x\\

\cos 4x &= 8 \cos^4 x – 8 \cos^2 x + 1 \\

\cos 5x &= 16 \cos^5 x – 20 \cos^3 x + 5 \cos x

\end{align}

and so on. (Formulas taken from here.)

Why does this matter? It means that we can find an explicit polynomial which has $\cos \frac{a\pi}{b}$ as a root. To see an example of this, let’s look at $\alpha = \cos \frac{2\pi}{5}$. Using the last formula above, we know that $$

\cos (2\pi)=\cos\left(5 \cdot \frac{2\pi}{5}\right)=16 \cos^5 \frac{2\pi}{5}- 20 \cos^3 \frac{2\pi}{5} + 5 \cos \frac{2\pi}{5}=16\alpha^5-20\alpha^3+5\alpha

$$

Since $\cos(2\pi)=1$, we know that $\alpha$ is a solution to the equation $16x^5-20x^3+5x=1$, or, to put it another way, a root of the polynomial $16x^5-20x^3+5x-1$. So, by definition, $\alpha$ is algebraic (i.e., not transcendental).

You can use multiple-angle formulas to do this in general. If you want to prove that $\beta=\cos\frac{a\pi}{b}$ is algebraic, you just need to use the formula for $\cos bx$. This will give you a polynomial expression for $\cos(a\pi)$ in terms of $\beta$; since we know that $\cos(a\pi)=\pm 1$, that’s enough to prove that $\beta$ is algebraic.

So we just need to prove this key fact, which can be done by induction. For the base case, notice that $\cos x$ and $\cos 2x=2\cos^2 x -1$ are both polynomial functions of $\cos x$. For the inductive step, we’ll suppose that $\cos nx$ and $\cos (n-1)x$ are polynomials, and use the identity given here for $\cos a + \cos b$, with $a=(n+1)x$, $b=(n-1)x$:

$$\cos(n+1)x + \cos(n-1)x=2\cos nx \cos x$$

This can be rearranged into the formula $\cos(n+1)x = 2 \cos x \cos nx – \cos(n-1)x$, which means that if $\cos nx$ and $\cos(n-1)x$ are polynomials in $\cos x$, then so is $\cos(n+1)x$. So the proof by induction is finished.

If you’re interested in knowing more details about this kind of polynomial formula for $\cos nx$ in terms of $\cos x$, a useful keyword to search on would be “Chebyshev polynomials.”

As a final note, we could try to do this proof directly for $\sin \frac{a\pi}{b}$, as there are multiple-angle formulas for $\sin$ which are similar to the ones for $\cos$. But it’d be a little bit trickier, because those formulas often involve both $\sin$ and $\cos$. For example, $\sin 2x=2\cos x \sin x$, and getting rid of the $\cos$ would involve taking a square root and doing a bunch of annoying case analysis involving signs.

**Attribution***Source : Link , Question Author : Cubbs , Answer Author : Micah*