Mathematics: Why is cos(nt) = (-1)^n true, when n could be any integer?

Why is cos(nt) = (-1)^n true, when n could be any

josephfoxworth

11.03.2022

It's not true for any , but it is for $t=\pi$ .

Start with n=0 (which is even). Then

$\cos(0\pi)=\cos0=1$

The cosine function is $2\pi[tex]-periodic, meaning that for any [tex]x$ we have

$\cos(x+2\pi)=\cos x$

Now, $2\pi=0+2\pi$ , so

$\cos(2\pi)=\cos(0+2\pi)=\cos0=1$

This is the basis for an argument via induction to show that $\cos(n\pi)=1$ whenever is even. Also, when is even we have (-1)^n=1 .

In a similar way, starting with the fact that

$\cos(1\pi)=\cos\pi=-1$

and that $\cos x$ is periodic, we have

$\cos(3\pi)=\cos(\pi+2\pi)=\cos\pi=-1$

and so for odd we have $\cos(n\pi)=-1=(-1)^n$ .

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unkno

26.06.2020

-1048

Step-by-step explanation:

To find a specific term in an arthimatic sequence you use this formula.

an=a1+(n−1)d

an, is the the number of the term youre trying to find.

a1, is the number of the first term in the sequence.

n, is the number of the term.

d, is the difference between terms.

when your sequence is put into this formula you get, -1048 = -3 + (96 - 1) * -11

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