Annette Arroyo

2021-08-17

Verify for any integer n: $\mathrm{cos}\frac{\left(2n-1\right)\pi }{2}=0$. Be sure to include an explanation of potential cases that may exist with the value of n.

SchulzD

This is question related trigonometry basic problem
$\mathrm{cos}\frac{\left(2n-1\right)\pi }{2}$
$\mathrm{cos}\left(\frac{2n}{2}-\frac{1}{2}\right)\pi$
$⇒\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)$
if n is even
then $\mathrm{cos}\left(n\pi -0\right)=\mathrm{cos}\theta$
then $\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)=\mathrm{cos}\left(\frac{\pi }{2}\right)=0$
if n is odd
then $\mathrm{cos}\left(n\pi -\theta \right)=-\mathrm{cos}\theta$
hence $\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)=-\mathrm{cos}\frac{\pi }{2}=0$
hence we can say that
$\mathrm{cos}\left(\frac{2n-1}{2}\right)\pi =0$ for all integral values of n" на "This is question related trigonometry basic problem
$\mathrm{cos}\frac{\left(2n-1\right)\pi }{2}$
$\mathrm{cos}\left(\frac{2n}{2}-\frac{1}{2}\right)\pi$
$⇒\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)$
if n is even
then $\mathrm{cos}\left(n\pi -0\right)=\mathrm{cos}\theta$
then $\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)=\mathrm{cos}\left(\frac{\pi }{2}\right)=0$
if n is odd
then $\mathrm{cos}\left(n\pi -\theta \right)=-\mathrm{cos}\theta$
hence $\mathrm{cos}\left(n\pi -\frac{\pi }{2}\right)=-\mathrm{cos}\frac{\pi }{2}=0$
hence we can say that
$\mathrm{cos}\left(\frac{2n-1}{2}\right)\pi =0$ for all integral values of n
Verify for some integral value of "n"
for n=1 $\mathrm{cos}\left(\pi -\frac{\pi }{2}\right)=\mathrm{cos}\left(\frac{\pi }{2}\right)=0$
for n=2 $\mathrm{cos}\left(2\pi -\frac{\pi }{2}\right)=\mathrm{cos}\left(\frac{3\pi }{2}\right)=0$
for n=3 $\mathrm{cos}\left(3\pi -\frac{\pi }{2}\right)=-\mathrm{cos}\left(\frac{\pi }{2}\right)=0$

Jeffrey Jordon