trigonometry - Why $\sin(n\pi) = 0$ and $\cos(n\pi)=(-1)^n ...
WEB$ => sin(n\pi) = \frac{exp(in\pi) - exp(-in\pi)}{2i}$ You can think of the positive exponential term as rotating counterclockwise, and the negative exponential term rotating clockwise. Since both exponential terms are of equal value (i.e. 1), but only moving in opposite directions, they cancel out at all points n
real analysis - Show that $\sin \dfrac{n \pi}{4}$ is divergent ...
WEBIn the case of your first sequence, by this fact all elements are equal either to $\sin 0$ or to $\sin \pi.$ For the second sequence, the same fact shows that all elements are equal to $\sin(\frac\pi2).$
trigonometry - Why is the sin of n times pi always 0 - Mathematics ...
WEBMar 22, 2016 · 1. quite naturally it is because of precision. By definition of the sine function,we have that sin(nπ) = 0 s i n (n π) = 0 for all integers n n, so the only thing that could affect your answer, since the sine function is defined correctly, is the precision. Of course, π π is irrational, hence you need a non-recurring non-terminating decimal ...
How is sin(npi/2) = (-1)^((n-1)/2) for odd n? - Physics Forums
WEBSep 19, 2009. In summary, the value of sin (npi/2) for odd values of n can be calculated using the formula (-1)^ ( (n-1)/2), which is derived from the properties of sine function and the relationship between sine and cosine functions. This formula holds true for odd values of n because npi/2 falls in the second or fourth quadrant of the unit ...
Prove $a_n = \\sin(n\\pi /2)$ does not converge to $0$
WEBApr 15, 2016 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
sequences and series - Limit of $\sin(\pi n)$ as $n$ approaches ...
WEBApr 19, 2017 · Tour Start here for a quick overview of the site
limit of n sin (pi/n) as n - Mathematics Stack Exchange
WEBx ≈ x. As n → ∞ n → ∞, π n → 0 π n → 0. Fact 2 implies that, for large n n, sin π n ≈ π n sin π n ≈ π n. Then. limn→∞(n ⋅ sin(π n)) = limn→∞(n ⋅ π n) = limn→∞ π = π lim n → ∞ (n ⋅ sin (π n)) = lim n → ∞ (n ⋅ π n) = lim n → ∞ π = π. Share. edited Nov 2, 2019 at 3:00. alphacapture ...
real analysis - Limit of $sin(n\pi/2)$ as $n$ goes to $\infty ...
WEBApr 13, 2023 · Exercise 3.2, Question 16 from Introduction to Real Analysis by Bartle & Sherbert: Show that $\lim_{n \to \infty} \sin \frac{n\pi}{2}$ doesn't exist.
$n\\in \\mathbb{N} :\\ \\ \\ \\ \\sin (n \\pi +x)=(-1)^n \\sin x
WEBStack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Integral of $\\sin(\\pi x) \\cos(n\\pi x)$ and $\\sin(\\pi x) \\sin(n ...
WEBSep 5, 2015 · $$ \frac{-\sin(\pi x) \cos(n \pi x)}{n \pi} + \frac{1}{n}\int_0^1 \cos(\pi x) \cos(n \pi x) \, dx$$ And it does look like this will end up with the original integral in it again after another iteration, but the algebra this is generating is already getting beyond me.