De moivres theorem

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January 28th 2009, 10:57 AM
djmccabie

De moivres theorem

Simplify the following

A) (cosx - isinx)^5
B) (sinx - icosx)^4

for A i get

cos5x - isin5x
I think this is right...?

not sure if B is done the same or not. any help?
January 28th 2009, 11:03 AM
Moo

Hello,

Quote:

Originally Posted by djmccabie

Simplify the following

A) (cosx - isinx)^5
B) (sinx - icosx)^4

for A i get

cos5x - isin5x
I think this is right...?

not sure if B is done the same or not. any help?

Here is what you can do :)

De Moivre's formula says : $(\cos(x)+i\sin(x))^n=\cos(nx)+i \sin(nx)$

Remember that the sine function is odd and the cosine function is even.
So $(\cos(x)-i\sin(x))^n=(\cos(-x)+i \sin(-x))^n=$ $\cos(-nx)+i \sin(-nx)=\cos(nx)-i \sin(nx)$ (Wink)
January 28th 2009, 11:05 AM
djmccabie

so did i do it right :S

is it different when cosine and sine are switched round?

thanks
January 28th 2009, 11:09 AM
Moo

Quote:

Originally Posted by djmccabie

so did i do it right :S

is it different when cosine and sine are switched round?

thanks

Yes, it's the correct result.

If they're switched... do you mean $(-\cos(x)+i \sin(x))^n$ ?
Then it's just $(-1)^n (\cos(x)-i\sin(x))^n=(-1)^n(\cos(nx)-i\sin(nx))$

:p
January 28th 2009, 11:21 AM
djmccabie

hi thanks for helping me. its actualyy (sinx - icosx) does this change things??
January 28th 2009, 11:25 AM
Moo

Quote:

Originally Posted by djmccabie

hi thanks for helping me. its actualyy (sinx - icosx) does this change things??

Yes :

$\sin(x)-i \cos(x)$
Multiply and divide by i :

$\frac 1i \left(i [\sin(x)-i\cos(x)]\right)=\frac 1i (\cos(x)+i \sin(x))$

So $(\sin(x)-i \cos(x))^n=\frac{1}{i^n} (\cos(nx)+i \sin(nx))$

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