Question

If $m, n$ and $p$ are integers, such that $\ell = \sin\frac{m\pi}{p}\sin\frac{n\pi}{p} + \sin\frac{2m\pi}{p}\sin\frac{2n\pi}{p} + \sin\frac{3m\pi}{p}\sin\frac{3n\pi}{p} + \dots + \sin\frac{(p-1)m\pi}{p}\sin\frac{(p-1)n\pi}{p}$, then $\ell$ is equal to

• A. -11, if $m=151, n=25, p=22$
• B. $\frac{p}{2}$, if $m-n$ is divisible by $2p$, and $m+n$ is NOT.
• C. 1, if $m=125, n=34, p=79$
• D. 1, if both $m+n$ nor $m-n$ is divisible by $2p$.

Answer 1

Answer: (A)

A

Answer 2

The given sum is $\ell = \sum_{k=1}^{p-1} \sin\left(\frac{km\pi}{p}\right) \sin\left(\frac{kn\pi}{p}\right)$ Using the product-to-sum identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$, we get: $\ell = \sum_{k=1}^{p-1} \frac{1}{2} \left[ \cos\left(\frac{k(m-n)\pi}{p}\right) - \cos\left(\frac{k(m+n)\pi}{p}\right) \right]$ $\ell = \frac{1}{2} \left[ \sum_{k=1}^{p-1} \cos\left(\frac{k(m-n)\pi}{p}\right) - \sum_{k=1}^{p-1} \cos\left(\frac{k(m+n)\pi}{p}\right) \right]$ Let $S(\theta) = \sum_{k=1}^{p-1} \cos\left(\frac{k\theta\pi}{p}\right)$.

We analyze $S(\theta)$:

1. If $\theta$ is a multiple of $2p$, $\theta = 2jp$ for some integer $j$. $S(\theta) = \sum_{k=1}^{p-1} \cos(2jk\pi) = \sum_{k=1}^{p-1} 1 = p-1$.

2. If $\theta$ is an even integer, but not a multiple of $2p$. Let $\theta = 2j$ where $j$ is not a multiple of $p$. Let $z = e^{i\frac{2j\pi}{p}}$. Then $z \neq 1$ and $z^p = e^{i2j\pi} = 1$. $S(\theta) = \text{Re}\left(\sum_{k=1}^{p-1} z^k\right) = \text{Re}\left(\frac{z(1-z^{p-1})}{1-z}\right)$. Since $z^p=1$, $\sum_{k=0}^{p-1} z^k = 0$. So $\sum_{k=1}^{p-1} z^k = -z^0 = -1$. $S(\theta) = \text{Re}(-1) = -1$.

3. If $\theta$ is an odd integer. Let $z = e^{i\frac{\theta\pi}{p}}$. Then $z^p = e^{i\theta\pi} = (e^{i\pi})^\theta = (-1)^\theta = -1$ (since $\theta$ is odd). $S(\theta) = \text{Re}\left(\sum_{k=1}^{p-1} z^k\right)$. We know $\sum_{k=0}^{p-1} z^k = \frac{1-z^p}{1-z} = \frac{1-(-1)}{1-z} = \frac{2}{1-z}$. $\sum_{k=1}^{p-1} z^k = \frac{2}{1-z} - 1$. $S(\theta) = \text{Re}\left(\frac{2}{1-z}\right) - 1$. Let $x = \frac{\theta\pi}{p}$. $1-z = 1 - \cos x - i\sin x$. $\text{Re}\left(\frac{2}{1-z}\right) = \text{Re}\left(\frac{2(1-\cos x + i\sin x)}{(1-\cos x)^2 + \sin^2 x}\right) = \text{Re}\left(\frac{2(1-\cos x + i\sin x)}{2-2\cos x}\right) = \text{Re}\left(1 + i\frac{\sin x}{1-\cos x}\right) = 1$. So, $S(\theta) = 1 - 1 = 0$.

Summary for $S(\theta)$:

• $S(\theta) = p-1$ if $\theta$ is a multiple of $2p$.
• $S(\theta) = -1$ if $\theta$ is an even integer not divisible by $2p$.
• $S(\theta) = 0$ if $\theta$ is an odd integer.

Evaluating the options:

(A) $m=151, n=25, p=22$. $m-n = 126$. $m+n = 176$. $p=22$, $2p=44$. $m-n = 126$ is even, not divisible by $44$. $S(126) = -1$. $m+n = 176 = 4 \times 44$, divisible by $2p$. $S(176) = p-1 = 22-1 = 21$. $\ell = \frac{1}{2}[S(126) - S(176)] = \frac{1}{2}[-1 - 21] = \frac{1}{2}[-22] = -11$. Option (A) is correct.

(B) If $m-n$ is divisible by $2p$, $S(m-n)=p-1$. If $m+n$ is not divisible by $2p$: Case 1: $m+n$ is odd. $S(m+n)=0$. $\ell = \frac{1}{2}[(p-1)-0] = \frac{p-1}{2} \neq \frac{p}{2}$. Case 2: $m+n$ is even, not divisible by $2p$. $S(m+n)=-1$. $\ell = \frac{1}{2}[(p-1)-(-1)] = \frac{1}{2}[p] = \frac{p}{2}$. Since the outcome can be different, option (B) is incorrect.

(C) $m=125, n=34, p=79$. $m-n = 91$ (odd). $S(91)=0$. $m+n = 159$ (odd). $S(159)=0$. $\ell = \frac{1}{2}[0-0] = 0 \neq 1$. Option (C) is incorrect.

(D) If $m+n$ and $m-n$ are not divisible by $2p$: Case 1: $m-n$ and $m+n$ are both odd. $S(m-n)=0$, $S(m+n)=0$. $\ell = \frac{1}{2}[0-0]=0$. Case 2: $m-n$ and $m+n$ are both even, not divisible by $2p$. $S(m-n)=-1$, $S(m+n)=-1$. $\ell = \frac{1}{2}[-1-(-1)]=0$. In both cases, $\ell=0 \neq 1$. Option (D) is incorrect.