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 [1].

\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

[1] E. Maor, Trigonometric Delights, Princeton University Press, 1998.

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: