A Proof Using Binomial Expansion
Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 18, 2025
On the page of Other Trigonometric Fundamental Formulae we proved the following result:
$$
\begin{eqnarray*}
\sin^2(x) + \cos^2(x)
&=& \left( \sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) \right)^2 \\
&=& \left( \sin^2\left(\frac{x}{4}\right) + \cos^2\left(\frac{x}{4}\right) \right)^4 \\
&\vdots& \\
&=& \left( \sin^2\left(\frac{x}{2^n}\right) + \cos^2\left(\frac{x}{2^n}\right) \right)^{2^n}.
\end{eqnarray*}
$$
If is easy to show that the limit of the above as \( n \) approaches \( \infty \) is 1
using L'Hôpital's Rule.
Please see here for the details.
As a result, the Pythagorean Identity holds.
Our question is:
Can we do the same without the use of
L'Hôpital's Rule?
This page provides you with a positive answer.
From the Binomial Theorem, we have the following:
\[ \left( \sin^2\left(\frac{x}{2^n}\right) + \cos^2\left(\frac{x}{2^n}\right) \right)^{2^n}
= \sum_{i=0}^{2^n} \frac{(2^n)!}{i!(2^n-i)!}
\left(\sin^2\left(\frac{x}{2^n}\right)\right)^i
\left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n-i} .\]
We have three cases to consider:
- If \( i = 0 \), we have a term without \( \sin() \):
\[ \left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n}
= \left[ \left(\cos\left(\frac{x}{2^n}\right)\right)^{2^n} \right]^2. \]
- If \( i = 2^n \), we have a term without \( \sin() \):
\[ \left(\sin^2\left(\frac{x}{2^n}\right)\right)^{2^n}
= \left[ \left(\sin\left(\frac{x}{2^n}\right)\right)^{2^n} \right]^2. \]
- If \( 0 < i < 2^n \), every terms includes \( \sin() \) and \( \cos() \):
\[ \sum_{i=0}^{2^n} \frac{(2^n)!}{i!(2^n-i)!}
\left(\sin^2\left(\frac{x}{2^n}\right)\right)^i
\left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n-i}. \]
We shall prove that as \( n \) approaches \( \infty \)
the first term (1) approaches 1,
the other terms ((2) and (3)) approaches 0.
Hence, the desired result is proved.
\( i = 0 \): Only \( \cos() \)
What we need to prove is the following:
\[ \lim_{n\rightarrow\infty} \left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n}
= \lim_{n\rightarrow\infty} \left[ \left(\cos\left(\frac{x}{2^n}\right)\right)^{2^n} \right]^2 = 1. \]
Let \( y \) be
\[ y = \left[ \left(\cos\left(\frac{x}{2^n}\right)\right)^{2^n} \right]^2. \]
Then, let \( k = \frac{1}{2^n} \) and after taking \( \ln() \) yields
\[ \ln(y) = 2\ln\left[ \cos(kx)^{\frac{1}{k}} \right] = 2\cdot\frac{\ln(\cos(kx))}{k} \]
Therefore, from what we know on the A Few Useful Results page,
we have
\[ \lim_{k\rightarrow 0} \ln(y) = \lim_{k\rightarrow 0} \left(2x\cdot\frac{\ln(\cos(kx))}{kx}\right) =
\lim_{kx\rightarrow 0} \left(2x\cdot\frac{\ln(\cos(kx))}{kx}\right) =
(2x)\lim_{kx\rightarrow 0} \left(\frac{\ln(\cos(kx))}{kx}\right) = 0. \]
Consequently, the following holds
\[ \lim_{n\rightarrow\infty} \left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n}
= \exp(\lim_{k\rightarrow 0} \ln(y)) = \exp(0) = 1. \]
\( i = 2^n \): Only \( \sin() \)
This is a rather easy case.
Note that if \( 1 > a > 0 \), then \( a > a^2 \).
Because \( 1 > \sin(x) > 0 \) for every positive \( x \),
we have the following
\[ 1 > \sin\left(\frac{x}{2^n}\right)
> \sin^2\left(\frac{x}{2^n}\right)
> \sin^4 \left(\frac{x}{2^n}\right)
> \cdots
> \sin^{2^n}\left(\frac{x}{2^n}\right)
> 0. \]
Hence, the following holds
\[ 0 = \lim_{n\rightarrow\infty} \sin\left(\frac{x}{2^n}\right)
> \lim_{n\rightarrow\infty} \sin^{2^n}\left(\frac{x}{2^n}\right). \]
and we have the desired result.
\( 0 < i < 2^n-1 \): Mixed \( \sin() \) and \( \cos() \) Terms
If \( 0 < i < 2^n-1 \), each term has both \( \sin() \) and \( \cos() \) as follows:
\[ \frac{(2^n)!}{i!(2^n-i)!}
\left(\sin^2\left(\frac{x}{2^n}\right)\right)^i
\left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n-i}. \]
We wish to prove that the above term approaches 0 as \( n \) approaches \( \infty \).
Because \( \cos(x) \leq 1 \),
the following holds:
$$
\begin{eqnarray*}
\underbrace{\frac{(2^n)(2^n-1)(2^n-2)\cdots(2^n-i+1)}{i\cdot (i-1)\cdot(i-2)\cdot 3\cdot 2\cdot 1}}_{\text{\( i \) terms}} \left(\sin^2\left(\frac{x}{2^n}\right)\right)^i
\left(\cos^2\left(\frac{x}{2^n}\right)\right)^{2^n-i}
&<&
\underbrace{\frac{(2^n)(2^n-1)(2^n-2)\cdots(2^n-i+1)}{i\cdot (i-1)\cdot(i-2)\cdot 3\cdot 2\cdot 1}}_{\text{\( i \) terms}}
\left(\sin^2\left(\frac{x}{2^n}\right)\right)^i \\
&=& \left(\frac{2^n}{i}\right)\left(\frac{2^n-1}{i-1}\right)\left(\frac{2^n-2}{i-2}\right)
\cdots \left(\frac{2^n-i+1}{1}\right) \left(\sin^2\left(\frac{x}{2^n}\right)\right)^i \\
&<& \underbrace{2^n\cdot2^n\cdots 2^n }_{\text{\( i \) terms}} \left(\sin^2\left(\frac{x}{2^n}\right)\right)^i \\
% &=& \left( 2^n \right)^i \left(\sin^2\left(\frac{x}{2^n}\right)\right)\right)^i \\
&=&\left( 2^n \left(\sin^2\left(\frac{x}{2^n}\right)\right) \right)^i \\
&=& \left[ \sin\left(\frac{x}{2^n}\right)\left( 2^n \sin\left(\frac{x}{2^n}\right) \right) \right]^i \\
&=& \sin^i\left(\frac{x}{2^n}\right)
\left( \frac{\sin\left(\frac{x}{2^n}\right)}{\left( \frac{1}{2^n}\right)} \right)^i.
\end{eqnarray*}
$$
We have shown that as \( n \) approaches \( \infty \) the first term approaches 0
while the second term approaches 1.
Please see A Few Useful Results for a proof of \( \lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1 \).
Note that this proof is independent of the Pythagorean Theorem and the Pythagorean Identity.
Consequently, we have the expected result.
Summary
Combining all three cases, we have
\[ \lim_{n\rightarrow\infty} \left( \sin^2\left(\frac{x}{2^n}\right) + \cos^2\left(\frac{x}{2^n}\right) \right)^{2^n} = 1,\]
and hence \( \sin^2(x) = \cos^2(x) = 1 \).
I first proved this result using
L'Hôpital's Rule in late August 2023.
Please see this essay (Agust 21, 2023) or this video (October 31, 2023).
Then, I challenged myself to see whether L'Hôpital's Rule is really needed.
A very complex and lengthy proof was obtained using the \( \epsilon-\delta \) notation.
However, I was very dis-satisfied
and have been trying to re-do the proof on and off.
I reached the idea of using the Binomial Theorem but stopped there as I have other
interesting things to do.
Before I re-picked up this approach,
I found that the use of a dense set in \( \left[ 0,\frac{1}{2}\pi \right] \),
an idea used decades ago for a similar work,
can yield a very interesting proof.
Please see The Use of a Dense Set page
for the details.
Finally, I decided to come back to the Binomial Theorem idea
and finished the proof here.
Just like the dense set approach,
this proof is rather long and perhaps an overkill.
However, it does work.
Doesn't it?
Hope you like it.
Go back to
Home Page.