Prove that for every natural number $n$, and for every real number $x \neq \frac{k\pi}{2^t}$ ($t=0,1, \dots, n$; $k$ any integer) \[ \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx} \]
Problem
Source: IMO 1966, Day 2, Problem 4
Tags: trigonometry, induction, algebra, Trigonometric Identities, IMO, IMO 1966
boxedexe
16.08.2007 02:07
Solution. By the double angle formula for cosine, $ \cos 2^{k}x=2\cos^{2}2^{k-1}x-1$. Dividing both sides by $ \sin 2^{k}x$ ($ \neq 0$ by the given values of $ x$), \[ \cot 2^{k}x=\cot 2^{k-1}x-\frac{1}{\sin 2^{k}x}\implies\frac{1}{\sin 2^{k}x}=\cot 2^{k-1}x-\cot 2^{k}x.\] Hence, \[ \sum_{k=1}^{n}\frac{1}{\sin 2^{k}x}=\sum_{k=1}^{n}{(\cot 2^{k-1}x-\cot 2^{k}x)}=\cot x-\cot 2^{n}x.\] And the conclusion follows. $ \Box$
Altheman
17.08.2007 18:47
I understand that the difficulties of old IMO problems are lower, but this is a bit ridiculous. Even if you don't see the telescoping solution, you can just use induction. It is a very nice problem though.
AwesomeToad
14.07.2011 21:43
Sorry to revive, but here's another solution.
Base case, $n=1$ gives $\frac{1}{\sin{2x}}=\cot{x}-\cot{2{x}}$, which is true: $\cot{x}-\cot{2x}=\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}=\frac{\cos^2{x}}{\sin{2x}}-\frac{\cos^2{x}-1}{\sin{2x}}=\frac{1}{\sin{2x}}$.
Assume that the identity holds for $n=k$. For $n=k+1$, we get
$\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\cdots+\frac{1}{\sin{2^k x}}+\frac{1}{\sin{2^{k+1}x}}$
$=\cot{x}-\cot{2^k{x}}+\frac{1}{\sin{2^{k+1}x}}$
$=\cot{x}-\frac{\cos{2^k x}}{\sin{2^k x}}+\frac{1}{2\sin{2^kx}\cos{2^kx}}$
$=\cot{x}-\frac{2(\cos{2^kx})^2}{2\sin{2^kx}\cos{2^kx}}+\frac{1}{2\sin{2^kx}\cos{2^kx}}$
$=\cot{x}-\frac{2(cos{2^kx})^2-1}{2\sin{2^kx}\cos{2^kx}}$
$=\cot{x}-\frac{\cos{2^{k+1}x}}{\sin{2^{k+1}x}}$
$=\cot{x}-\cot{2^{k+1}}$
So the identity holds for $n=k+1$, completing the induction.
bobthegod78
18.03.2022 16:46
Also used induction.
We can use induction. The base case is $n=1.$ We get $\frac{1}{\sin(2x)} = \cot x - \cot(2x)$. Let's simplify the RHS first. To make it easier, we can write this in terms of $\tan$ and use the double angle formula to get $\frac{1}{\tan x} - \frac{1-\tan^2 x}{2\tan x} = \frac{\tan^2 x +1}{2\tan x}.$ Writing this in terms of $\sin$ and $\cos$, we get $$\frac{\frac{\sin^2 x}{\cos^2 x}+1}{\frac{2\sin x}{\cos x}} = \frac{\sin^2 x + \cos^2 x}{2\sin x \cos x } = \frac{1}{2 \sin x \cos x}.$$But $\frac{1}{\sin(2x)} = \frac{1}{2\sin x \cos x}$. We are done. Now for the inductive step. We need to show that $\frac{1}{\sin{2^{n+1}x}} = \cot(2^n x) - \cot(2^{n+1}x).$ Again, we first work with the RHS. We can write this as (using double angle formula) $$\frac{1}{\tan(2^n x)} - \frac{1-\tan^2 (2^n x)}{2\tan(2^n x)} = \frac{\tan^2 (2^n x)+1}{2\tan(2^n x)}.$$Again, we can write this in terms of $\sin$ and $\cos$. We get $$\frac{\frac{\sin^2 (2^n x)}{\cos^2 (2^n x)}+1}{\frac{2\sin(2^n x)}{\cos(2^n x)}} = \frac{\sin^2 (2^n x) + \cos^2(2^n x)}{2\sin (2^n x) \cos(2^n x)} = \frac{1}{2\sin (2^n x) \cos(2^n x)}.$$But $\frac{1}{\sin(2^{n+1} x)} = \frac{1}{2\sin (2^n x) \cos (2^n x)}.$ We are done with the inductive step, so we are done.
hanulyeongsam
19.10.2024 10:58
It is easy to prove that {1/tan(x)}-{1/tan(2x)}=1/sin(2x)
{1/tan(2x)} =[1/{2*tan(x)/(1-tan^2(x))}]=(1-tan^2(x))/2tan(x)
1/tan(x)-1/tan(2x)=(1/2 )*(tan(x)+1/tan(x))
=1/sin(2x)
cot(x)-cot(2^n*x)