IMO 1966 Problem 4
The identity resembles a telescoping sum.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m33s
Problem
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) $$ \frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx} $$
Exploration
The identity resembles a telescoping sum. The summands contain reciprocals of $\sin(2^k x)$, while the right-hand side contains cotangents at consecutive powers of $2$. The first task is to examine small values of $n$.
For $n=1$, the statement becomes
$$\frac1{\sin 2x}=\cot x-\cot 2x.$$
Using
$$\cot x-\cot 2x = \frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x},$$
and bringing the terms to a common denominator,
$$\cot x-\cot 2x = \frac{\cos x\sin 2x-\sin x\cos 2x} {\sin x\sin 2x}.$$
Since
$$\cos x\sin 2x-\sin x\cos 2x = \sin(2x-x) = \sin x,$$
we obtain
$$\cot x-\cot 2x = \frac1{\sin 2x},$$
which matches the claim.
For $n=2$,
$$\frac1{\sin 2x}+\frac1{\sin 4x} = (\cot x-\cot 2x)+(\cot 2x-\cot 4x) = \cot x-\cot 4x.$$
The cancellation suggests that the general proof should come from an identity of the form
$$\frac1{\sin 2y}=\cot y-\cot 2y.$$
Then substituting $y=2^{k-1}x$ would give
$$\frac1{\sin(2^k x)} = \cot(2^{k-1}x)-\cot(2^k x),$$
and summing over $k$ would telescope.
A possible hidden difficulty is the validity of all denominators. The problem excludes numbers of the form $k\pi/2^t$ for $t=0,1,\dots,n$. This must guarantee that every sine appearing in the argument is nonzero, so that each cotangent and reciprocal sine is defined.
The proof should therefore consist of two ingredients. First, establish the elementary identity
$$\frac1{\sin 2y}=\cot y-\cot 2y.$$
Second, apply it successively with $y=x,2x,\dots,2^{n-1}x$ and telescope.
Problem Understanding
The problem asks us to prove an identity involving a finite trigonometric sum. For every natural number $n$ and every real number $x$ for which none of the denominators vanish, we must show that
$$\sum_{k=1}^{n}\frac1{\sin(2^k x)} = \cot x-\cot(2^n x).$$
This is a Type B problem. The statement is already specified, and the task is to establish its validity.
The mathematical objects involved are the trigonometric functions $\sin$ and $\cot$. The exclusion
$$x\neq \frac{k\pi}{2^t}, \qquad t=0,1,\dots,n,$$
ensures that every expression occurring in the identity is defined.
The core difficulty is recognizing the hidden telescoping structure. A direct attempt to combine the reciprocals of the sines does not reveal any obvious simplification. The key observation is that each reciprocal sine can be rewritten as the difference of two cotangents, after which consecutive terms cancel.
Proof Architecture
The proof uses one lemma and then a telescoping summation.
Lemma 1 states that for every real number $y$ such that $\sin y\neq0$ and $\sin 2y\neq0$,
$$\frac1{\sin 2y}=\cot y-\cot 2y.$$
Its proof comes from expressing the cotangents as ratios of sine and cosine and then applying the subtraction formula for sine.
After establishing the lemma, we substitute $y=2^{k-1}x$ for each $k=1,\dots,n$. This rewrites every summand as
$$\cot(2^{k-1}x)-\cot(2^k x).$$
Summing these identities produces a telescoping series in which all intermediate cotangent terms cancel.
The most delicate point is verifying that all denominators are nonzero. The hypothesis on $x$ must be used to ensure that $\sin(2^m x)\neq0$ for every relevant exponent $m$.
Solution
We first verify that all expressions in the identity are defined.
For each integer $m$ with $0\le m\le n$, suppose that $\sin(2^m x)=0$. Then
$$2^m x=k\pi$$
for some integer $k$, and hence
$$x=\frac{k\pi}{2^m}.$$
This contradicts the hypothesis. Consequently,
$$\sin(2^m x)\neq0 \qquad (0\le m\le n).$$
Therefore every reciprocal sine and every cotangent appearing below is defined.
Lemma 1
For every real number $y$ satisfying $\sin y\neq0$ and $\sin 2y\neq0$,
$$\frac1{\sin 2y} = \cot y-\cot 2y.$$
Proof
Using the definition of cotangent,
$$\cot y-\cot 2y = \frac{\cos y}{\sin y} - \frac{\cos 2y}{\sin 2y}.$$
Combining the fractions,
$$\cot y-\cot 2y = \frac{\cos y\sin 2y-\sin y\cos 2y} {\sin y\sin 2y}.$$
The sine subtraction formula,
$$\sin(A-B)=\sin A\cos B-\cos A\sin B,$$
with $A=2y$ and $B=y$, gives
$$\cos y\sin 2y-\sin y\cos 2y = \sin(2y-y) = \sin y.$$
Substituting this into the preceding expression yields
$$\cot y-\cot 2y = \frac{\sin y}{\sin y\sin 2y} = \frac1{\sin 2y}.$$
The lemma is proved. ∎
Certification: this lemma converts each reciprocal sine into a difference of cotangents, and omitting the common-denominator computation would leave the crucial identity unproved.
We now apply Lemma 1 with
$$y=2^{k-1}x, \qquad k=1,2,\dots,n.$$
Since $\sin(2^{k-1}x)\neq0$ and $\sin(2^k x)\neq0$, the lemma gives
$$\frac1{\sin(2^k x)} = \cot(2^{k-1}x)-\cot(2^k x).$$
Summing these identities from $k=1$ to $k=n$,
$$\sum_{k=1}^{n}\frac1{\sin(2^k x)} = \sum_{k=1}^{n} \bigl( \cot(2^{k-1}x)-\cot(2^k x) \bigr).$$
Writing out the right-hand side,
$$\begin{aligned} &(\cot x-\cot 2x) +(\cot 2x-\cot 4x) +\cdots \ &\qquad +(\cot(2^{n-1}x)-\cot(2^n x)). \end{aligned}$$
Every intermediate term cancels with its successor, leaving
$$\sum_{k=1}^{n}\frac1{\sin(2^k x)} = \cot x-\cot(2^n x).$$
This is exactly the required identity.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$\frac1{\sin 2y} = \cot y-\cot 2y.$$
Re-deriving it independently, we compute
$$\cot y-\cot 2y = \frac{\cos y\sin 2y-\sin y\cos 2y} {\sin y\sin 2y}.$$
The numerator equals $\sin(2y-y)=\sin y$, hence the fraction becomes
$$\frac{\sin y}{\sin y\sin 2y} = \frac1{\sin 2y}.$$
A careless argument might incorrectly use $\sin 2y=2\sin y\cos y$ and then manipulate reciprocals without first checking that $\sin y\neq0$.
The second delicate step is the domain verification. If $\sin(2^m x)=0$, then $2^m x=k\pi$, so
$$x=\frac{k\pi}{2^m}.$$
Since the hypothesis excludes precisely these values for every $m=0,1,\dots,n$, every denominator is nonzero. Ignoring this check would permit undefined terms such as $\cot(2^m x)$.
The third delicate step is the telescoping cancellation. After substitution, the sum becomes
$$(\cot x-\cot 2x)+(\cot 2x-\cot 4x)+\cdots.$$
Each negative term $-\cot(2^j x)$ is paired with the positive term $+\cot(2^j x)$ from the next summand. The only surviving terms are the first positive term and the last negative term. Missing even one index shift would produce an incorrect endpoint.
Alternative Approaches
A different proof begins with the double-angle identity
$$\sin 2y=2\sin y\cos y.$$
Then
$$\frac1{\sin 2y} = \frac1{2\sin y\cos y}.$$
Using
$$\cot y=\frac{\cos y}{\sin y}, \qquad \tan y=\frac{\sin y}{\cos y},$$
one obtains
$$\frac1{\sin 2y} = \frac12(\cot y+\tan y).$$
Replacing $\tan y$ by $\cot y-2\cot 2y$, which follows from the double-angle formula for cotangent, yields again
$$\frac1{\sin 2y} = \cot y-\cot 2y.$$
The remainder is the same telescoping argument.
The main approach is preferable because it derives the key identity directly from the sine subtraction formula. It uses only one elementary lemma and exposes the telescoping structure immediately.