---
title: "Period of cos(πn/5)·sin(πn/3)"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:energy-power-period-cos-sin-product"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-period-cos-sin-product
site: BCT Engineering
---

# Period of cos(πn/5)·sin(πn/3)

Signals and Systems · Chapter 1: Signals

- Asked in: 2081 Asoj, 6 marks
- Video: [Periodic Discrete Time Signals (Solved Problems)](https://www.youtube.com/watch?v=dD_67ZmfPTc) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/energy-power-period-cos-sin-product
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Define energy and power signal. State whether the given signal is periodic or not? If signal is periodic, find its fundamental period. x[n] = cos(πn/5) sin(πn/3)

## Short answer

Energy signal has finite energy, zero power; power signal has finite power, infinite energy. Using product-to-sum, x[n]=½ sin(8πn/15)+½ sin(2πn/15). Both frequencies are rational (4/15 and 1/15 cycles/sample), so the signal is periodic. The LCM of periods (15,15) is 15, so fundamental period N=15; it is a power signal.

## Answer

**Energy signal:** $0 < E < \infty$, $P = 0$, where $E = \sum_{n}|x[n]|^2$.
**Power signal:** $0 < P < \infty$, $E = \infty$, where $P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{-N}^{N}|x[n]|^2$. Periodic sequences are power signals.

#### Periodicity of $x[n] = \cos\left(\frac{\pi n}{5}\right)\sin\left(\frac{\pi n}{3}\right)$

Use $\cos A\sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$:

$$
\begin{aligned}
x[n] &= \frac{1}{2}\left[\sin\left(\frac{\pi}{5}+\frac{\pi}{3}\right)n - \sin\left(\frac{\pi}{5}-\frac{\pi}{3}\right)n\right] \\
&= \frac{1}{2}\sin\left(\frac{8\pi}{15}n\right) + \frac{1}{2}\sin\left(\frac{2\pi}{15}n\right)
\end{aligned}
$$

- $\omega_1 = 8\pi/15$: $\omega_1/2\pi = 4/15$ (rational), $N_1 = 15$.
- $\omega_2 = 2\pi/15$: $\omega_2/2\pi = 1/15$ (rational), $N_2 = 15$.

$$
N = \text{LCM}(15, 15) = 15
$$

**Check:** $x[n+15] = \cos(\frac{\pi n}{5}+3\pi)\sin(\frac{\pi n}{3}+5\pi) = (-\cos\frac{\pi n}{5})(-\sin\frac{\pi n}{3}) = x[n]$.

(Taking the LCM of the periods of the two factors, 10 and 6, gives 30, which is a period but not the fundamental one.)

Since it is periodic, it is a **power signal**: $P = \frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\right) = \frac{1}{4}$.

**Answer:** periodic, fundamental period $N = 15$ samples ($\omega_0 = 2\pi/15$ rad/sample).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. cos(pi n/5) sin(pi n/3) is rewritten as a sum of sines using:
   A. (1/2)[sin(A+B) - sin(A-B)]
   B. cos A sin B = sin A cos B
   C. cos A sin B = sin(AB)
   D. cos A sin B = (1/2)[cos(A+B) + cos(A-B)]
   Answer: A. cos A sin B = (1/2)[sin(A+B) - sin(A-B)]. (Key point: Product-to-sum step)

2. The sum form contains frequencies:
   A. 8 pi/15 and 2 pi/15
   B. pi/5 and pi/3
   C. pi/15 and pi/8
   D. 8 pi/15 only
   Answer: A. A+B = 8 pi/15 and A-B = -2 pi/15, giving components at 8 pi/15 and 2 pi/15. (Key point: Resulting frequencies)

3. The fundamental period of x[n] is:
   A. 30
   B. 15
   C. 10
   D. 6
   Answer: B. Both components have period 15 (4/15 and 1/15 in lowest terms), so the LCM is 15. (Key point: Fundamental period)

4. Why is 30 (the LCM of 10 and 6) not the answer?
   A. It is not a period
   B. It is a period but not the smallest
   C. It is not an integer
   D. It gives a different signal
   Answer: B. The product repeats sooner than the factors' LCM because of sign cancellation; 15 is the fundamental period. (Key point: LCM pitfall)

5. Because x[n] is periodic, it is:
   A. An energy signal
   B. A power signal
   C. Neither
   D. Both
   Answer: B. Periodic sequences have infinite energy and finite power. (Key point: Signal class)

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