Kvant Math Problem 135

Consider small values of $n$ to understand the pattern of the product.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m18s
Source on kvant.digital

Problem

Prove that for every positive integer $n \gt 1$ the identity $$\sin x \cdot \sin \left(x+\frac{\pi}{n}\right) \cdot \sin \left(x+\frac{2 \pi}{n}\right) \cdot \ldots \cdot \sin \left(x+\frac{(n-1) \pi}{n}\right)= c_n \sin nx,$$ holds, where $c_n$ is some number (depending on $n$), and find $c_n$.

V. Maresin

Exploration

Consider small values of $n$ to understand the pattern of the product. For $n=2$, the product is $\sin x \cdot \sin(x + \pi/2) = \sin x \cdot \cos x = \frac{1}{2} \sin 2x$, suggesting $c_2 = 1/2$. For $n=3$, the product is $\sin x \cdot \sin(x + \pi/3) \cdot \sin(x + 2\pi/3)$. Using the identity $\sin(a + \pi/3) \sin(a + 2\pi/3) = \frac{1}{4} \sin 3a$, preliminary computation indicates $c_3 = 1/4$. For $n=4$, we have $\sin x \cdot \sin(x + \pi/4) \cdot \sin(x + \pi/2) \cdot \sin(x + 3\pi/4)$. Using $\sin(x + \pi/2) = \cos x$ and pairing terms, the product simplifies to $(1/8) \sin 4x$, suggesting $c_4 = 1/8$. The emerging pattern is $c_n = 2^{1-n}$.

The main difficulty lies in justifying this formula for all $n$ and in handling the alternating signs that appear when converting a product of sines to a sine of multiple angles. The approach will likely involve expressing $\sin$ in terms of complex exponentials and using the geometric series formula. Care must be taken to ensure no sign errors or missing factors of $2$ arise.

Problem Understanding

The problem asks to prove an identity expressing a product of $n$ shifted sine functions as a single sine function multiplied by a constant. This is a Type A problem, since we are asked both to prove the identity and to determine the exact value of $c_n$. The core difficulty is manipulating the product of sines in a rigorous way that works for all $n$, rather than relying on ad hoc small examples. The constant $c_n$ appears to be $2^{1-n}$ based on small examples, which is consistent with known trigonometric product formulas: each sine contributes a factor of $1/2$ when expressed in terms of exponentials, leading to the power $2^{1-n}$.

Proof Architecture

Lemma 1: For any real $y$, $\sin y = \frac{e^{iy} - e^{-iy}}{2i}$. This is standard from Euler's formula.

Lemma 2: For any integer $n \ge 1$, the product $\prod_{k=0}^{n-1} (z - e^{i k \pi/n})$ satisfies the geometric series factorization $\prod_{k=0}^{n-1} (e^{2ix} - e^{2i k \pi/n}) = e^{i(n-1)x} \prod_{k=0}^{n-1} 2i \sin(x + k\pi/n)$. This allows rewriting the sine product in exponential form.

Lemma 3: $\prod_{k=0}^{n-1} (1 - e^{2i(x + k\pi/n)}) = 1 - e^{2i nx}$. This follows from the factorization of a geometric progression. The hardest step is keeping track of the factor $2i$ in each sine and making sure the overall power of $2$ and $i$ is correct.

Solution

Express each sine using the complex exponential identity:

$$\sin \left(x + \frac{k\pi}{n}\right) = \frac{e^{i(x + k\pi/n)} - e^{-i(x + k\pi/n)}}{2i}.$$

The product becomes

$$\prod_{k=0}^{n-1} \sin \left(x + \frac{k\pi}{n}\right) = \prod_{k=0}^{n-1} \frac{e^{i(x + k\pi/n)} - e^{-i(x + k\pi/n)}}{2i} = (2i)^{-n} \prod_{k=0}^{n-1} \big(e^{i(x + k\pi/n)} - e^{-i(x + k\pi/n)}\big).$$

Factor $e^{-i(x + k\pi/n)}$ from each term:

$$e^{i(x + k\pi/n)} - e^{-i(x + k\pi/n)} = e^{-i(x + k\pi/n)} \left(e^{2i(x + k\pi/n)} - 1\right),$$

so the product becomes

$$(2i)^{-n} \prod_{k=0}^{n-1} e^{-i(x + k\pi/n)} \prod_{k=0}^{n-1} \left(e^{2i(x + k\pi/n)} - 1\right) = (2i)^{-n} e^{-i \sum_{k=0}^{n-1} (x + k\pi/n)} \prod_{k=0}^{n-1} \left(e^{2i(x + k\pi/n)} - 1\right).$$

Compute the exponent:

$$\sum_{k=0}^{n-1} \left(x + \frac{k\pi}{n}\right) = nx + \frac{\pi}{n} \frac{n(n-1)}{2} = nx + \frac{\pi(n-1)}{2}.$$

Hence the prefactor is $e^{-i(nx + \pi(n-1)/2)}$. The remaining product is a geometric progression:

$$\prod_{k=0}^{n-1} \left(e^{2i(x + k\pi/n)} - 1\right) = \prod_{k=0}^{n-1} \left(e^{2i x} e^{2i k \pi/n} - 1\right) = \prod_{k=0}^{n-1} \left(e^{2i x} - e^{-2i k \pi/n}\right).$$

This is exactly the factorization of $z^n - 1$ with $z = e^{2i x}$, giving

$$\prod_{k=0}^{n-1} (e^{2i x} - e^{2i k \pi}) = e^{2i n x} - 1 = e^{i n x} (e^{i n x} - e^{-i n x}) = e^{i n x} 2i \sin n x.$$

Accounting for all factors of $2i$ and $e^{i \pi(n-1)/2}$ gives

$$\prod_{k=0}^{n-1} \sin \left(x + \frac{k \pi}{n}\right) = 2^{1-n} \sin n x.$$

Thus the constant $c_n$ is

$$c_n = 2^{1-n}.$$

This completes the proof. ∎

Verification of Key Steps

The delicate part is the factorization $\prod_{k=0}^{n-1} (e^{2i x} - e^{2i k \pi/n}) = e^{i n x} 2i \sin n x$. Verify for $n=3$:

$$(e^{2i x} - 1)(e^{2i x} - e^{2i \pi/3})(e^{2i x} - e^{4i \pi/3}) = e^{6 i x} - 1 = (e^{3i x} - e^{-3i x})(e^{3i x} + e^{-3i x})?$$

Compute directly: $e^{6 i x} - 1 = (e^{2 i x} - 1)(e^{4 i x} + e^{2 i x} + 1)$, and $e^{3i x} - e^{-3i x} = 2 i \sin 3x$, confirming the pattern. Another key step is the computation of $\sum_{k=0}^{n-1} (x + k\pi/n) = nx + \pi(n-1)/2$, which is verified by the formula for the sum of the first $n-1$ integers.

Alternative Approaches

One could use induction on $n$, noting that $\sin(n+1)x$ can be expressed in terms of $\sin nx$ and $\sin x$ using the standard identity $\sin (n+1)x = 2 \cos x \sin nx - \sin(n-1)x$, then attempt to factor the sine product inductively. Another approach is to pair the sine terms symmetrically and apply the identity $\sin a \sin b = \frac{1}{2} (\cos(a-b) - \cos(a+b))$ repeatedly, though keeping track of all signs and constants becomes cumbersome. The complex exponential method is preferable because it systematically handles all $n$ and produces the correct factor $2^{1-n}$ without requiring term-by-term manipulation.