Verify for any integer n: \cos\frac{(2n-1)\pi}{2}=0. Be sure to include a

Annette Arroyo 2021-08-17 Answered
Verify for any integer n: \(\displaystyle{\cos{{\frac{{{\left({2}{n}-{1}\right)}\pi}}{{{2}}}}}}={0}\). Be sure to include an explanation of potential cases that may exist with the value of n.

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SchulzD
Answered 2021-08-18 Author has 17718 answers
This is question related trigonometry basic problem
\(\displaystyle{\cos{{\frac{{{\left({2}{n}-{1}\right)}\pi}}{{{2}}}}}}\)
\(\displaystyle{\cos{{\left({\frac{{{2}{n}}}{{{2}}}}-{\frac{{{1}}}{{{2}}}}\right)}}}\pi\)
\(\displaystyle\Rightarrow{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}\)
if n is even
then \(\displaystyle{\cos{{\left({n}\pi-{0}\right)}}}={\cos{\theta}}\)
then \(\displaystyle{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}={\cos{{\left({\frac{{\pi}}{{{2}}}}\right)}}}={0}\)
if n is odd
then \(\displaystyle{\cos{{\left({n}\pi-\theta\right)}}}=-{\cos{\theta}}\)
hence \(\displaystyle{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}=-{\cos{{\frac{{\pi}}{{{2}}}}}}={0}\)
hence we can say that
\(\displaystyle{\cos{{\left({\frac{{{2}{n}-{1}}}{{{2}}}}\right)}}}\pi={0}\) for all integral values of 'n'" на "This is question related trigonometry basic problem
\(\displaystyle{\cos{{\frac{{{\left({2}{n}-{1}\right)}\pi}}{{{2}}}}}}\)
\(\displaystyle{\cos{{\left({\frac{{{2}{n}}}{{{2}}}}-{\frac{{{1}}}{{{2}}}}\right)}}}\pi\)
\(\displaystyle\Rightarrow{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}\)
if n is even
then \(\displaystyle{\cos{{\left({n}\pi-{0}\right)}}}={\cos{\theta}}\)
then \(\displaystyle{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}={\cos{{\left({\frac{{\pi}}{{{2}}}}\right)}}}={0}\)
if n is odd
then \(\displaystyle{\cos{{\left({n}\pi-\theta\right)}}}=-{\cos{\theta}}\)
hence \(\displaystyle{\cos{{\left({n}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}=-{\cos{{\frac{{\pi}}{{{2}}}}}}={0}\)
hence we can say that
\(\displaystyle{\cos{{\left({\frac{{{2}{n}-{1}}}{{{2}}}}\right)}}}\pi={0}\) for all integral values of 'n'
Verify for some integral value of "n"
for n=1 \(\displaystyle{\cos{{\left(\pi-{\frac{{\pi}}{{{2}}}}\right)}}}={\cos{{\left({\frac{{\pi}}{{{2}}}}\right)}}}={0}\)
for n=2 \(\displaystyle{\cos{{\left({2}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}={\cos{{\left({\frac{{{3}\pi}}{{{2}}}}\right)}}}={0}\)
for n=3 \(\displaystyle{\cos{{\left({3}\pi-{\frac{{\pi}}{{{2}}}}\right)}}}=-{\cos{{\left({\frac{{\pi}}{{{2}}}}\right)}}}={0}\)
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Answered 2021-12-10 Author has 10511 answers

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