# Chaitanya's Random Pages

## May 25, 2013

### Sine of half angles

Filed under: mathematics — ckrao @ 8:34 am

I found the following cool-looking formulas in . $\displaystyle \sin \frac{45^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}}{2}$
(n + 1 nested roots) $\displaystyle \sin \frac{15^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}}{2}$
(n + 2 nested roots) $\displaystyle \sin \frac{18^{\circ}}{2^n} = \frac{\sqrt{8 - 2\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}}{4}$
(n + 2 nested roots)

To prove them, we use the following form double angle formula for cosine. $\displaystyle 4 \sin^2 x = 2(1 - \cos 2x) = 2 - \sqrt{4 - 4 \sin^2 2x }\quad \quad (1)$

Now we evaluate this for different values of $x$. Firstly $\sin^2 45^{\circ} = 1/2$ implying from (1) that $4 \sin^2 22.5^{\circ} = 2 - \sqrt{2}.\quad \quad \quad \quad (2)$

Secondly, \begin{aligned}4 \sin^2 15^{\circ} &= 2(1 - \cos 30^{\circ})\\ &= 2 - \sqrt{3},\end{aligned}

so that from (1), $\displaystyle 4 \sin^2 7.5^{\circ} = 2 - \sqrt{4 - (2 - \sqrt{3}} = 2 - \sqrt{2 + \sqrt{3}}. \quad \quad (3)$

To find $\sin 18^{\circ}$ one could apply Ptolemy’s theorem to four points of a regular pentagon or alternatively set $y = 36^{\circ}$ and write \begin{aligned} \cos 3y &= -\cos (180^{\circ} - 3y)\\&= - \cos 2y.\end{aligned}

Using the double and triple angle formulae for cosine this becomes \begin{aligned} 4 \cos^3 y - 3\cos y = 1 - 2\cos^2 y \\ \hbox{i.e.}\ 4\cos^3 y + 2\cos^2 y - 3\cos y - 1 &= 0\\ \hbox{i.e.} \ (\cos y + 1) (4 \cos^2 y - 2 \cos y - 1) &= 0.\end{aligned}

From this the only valid (positive) solution is $\cos y = (2 + 2\sqrt{5})/8$, so from (1) $16 \sin^2 18^{\circ} = 8(1 - \cos 36^{\circ}) = 6 - 2\sqrt{5}.$

Hence applying (1) again, $\displaystyle 16 \sin^2 9^{\circ} = 2(4 - \sqrt{16 - 16 \sin^2 18^{\circ} }) = 8 - 2\sqrt{10 + 2\sqrt{5}}.\quad \quad (4)$

Now we set up a hypothesis for mathematical induction. Assume that $\displaystyle 4 \sin^2 \frac{45^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}$ $\displaystyle 4 \sin^2 \frac{15^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}$ $\displaystyle 16 \sin^2 \frac{18^{\circ}}{2^k} = 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}$

Note that these are equivalent to our initial three formulas. In (2)-(4) we have seen that these results are true for $k = 1$. Then using (1) and the inductive hypothesis, \begin{aligned} 4 \sin^2 \frac{45^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{45^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k+1 nested roots}}\end{aligned}

Similarly, \begin{aligned} 4 \sin^2 \frac{15^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{15^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+2 nested roots}}\end{aligned}

and \begin{aligned} 16 \sin^2 \frac{18^{\circ}}{2^{k+1}} &= 8 - 2\sqrt{16 - 16 \sin^2 \frac{18^{\circ}}{2^{k}}}\\&= 8 - 2\sqrt{16 - \left(8 - 2\underbrace{ \sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+2 nested roots}}.\end{aligned}

In each case we have shown that if the result is true for $n = k$, it is true for $n = k+1$. By the principle of mathematical induction, the result is true for $n = 1,2, \ldots$ and we are done.

#### Reference

 E. Maor, Trigonometric Delights, Princeton University Press, 1998.