## 6 Aug 2011

### A curious identity

I was trying to do a question the other day when I found an interesting identity that is a corollary of the double angle formulae. I've searched online for its name, but it might be so trivial that it doesn't have one. Instead of revealing the identity, I'll let you figure it out for yourself:
$\frac{1}{2^0}\cos(x_1) = \cos(x_1)$
$\frac{1}{2^1}\left(\cos(x_1+x_2)+\cos(x_1-x_2)\right)=\cos(x_1)\cos(x_2)$
$\frac{1}{2^{2}}\left(\begin{array}{c} \phantom{+}\cos(x_{1}+x_{2}+x_{3})\\ +\cos(x_{1}+x_{2}-x_{3})\\ +\cos(x_{1}-x_{2}+x_{3})\\ +\cos(x_{1}-x_{2}-x_{3})\end{array}\right)=\cos(x_{1})\cos(x_{2})\cos(x_{3})$
Do you see the pattern?
Can you prove the pattern (Hint: Try induction)?
Are there other similar identities involving sines and combinations of sines and cosines?

For those who are curious, the question that led me to this identity was:
Determine the value of $\cos\left(\frac{\pi}{7}\right)\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right).$

There is a simple way to do this question, and a hard way. Sadly the identity I found led to the hard way.

#### 1 comment:

1. That is a very very interesting identity! Good job in discovering it! Induction works, but I am curious whether or not there is some deep combinatorial meaning behind that identity (since it involves the term 2^n).

And for the cos(pi/7)*cos(2pi/7)*cos(4pi/7), I guess the "easy way" would be to write it as real part of complex exponential.