De Moivre's Theorem (1 Viewer)

[NubMuncher]

Let z = cis x.

Show that z^n - z^-n = 2 cos nx and z^n - z^-n = 2i sin nx.

I get up to here :
cos(nx) - cos(-nx) + isin(nx) - isin(-nx)

So does cos(nx) = -cos(-nx) and isin(nx) = -isin(-nx) ? This would lead to the correct answer however im not sure if this is allowable.

 

[ohexploitable]

Show that z^n - z^-n = 2 cos nx and z^n - z^-n = 2i sin nx.

Is there a mistake there?

 

[NubMuncher]

that is what is written in the book, however now looking at the worked solutions, there may be a typo in the book.

The worked solutions say z^n + z^-n = 2cos nx and z^n - z^-n = 2isin nx.

If you have the Cambridge 4u book the question is ex 2.4 Q5

Confused XD

 

[ohexploitable]

Yeah, there's a typo, you wrote a minus instead of a plus.

 

[ohexploitable]

 

[NubMuncher]

Why does cos -nx = cos nx and isin -nx = -isin nx ?

 

[ohexploitable]

Why does cos -nx = cos nx and isin -nx = -isin nx ?

Trig identities, cos(-x)=cos(x), sin(-x)=-sin(x).

 

[NubMuncher]

LOL fail. Been working on 4u for too long that i've forgotten 2u xD

Thanks