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?

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- Jan 28th 2009, 10:57 AMdjmccabieDe 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? - Jan 28th 2009, 11:03 AMMoo
Hello,

Here is what you can do :)

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

Remember that the sine function is odd and the cosine function is even.

So $\displaystyle (\cos(x)-i\sin(x))^n=(\cos(-x)+i \sin(-x))^n=$ $\displaystyle \cos(-nx)+i \sin(-nx)=\cos(nx)-i \sin(nx)$ (Wink) - Jan 28th 2009, 11:05 AMdjmccabie
so did i do it right :S

is it different when cosine and sine are switched round?

thanks - Jan 28th 2009, 11:09 AMMoo
- Jan 28th 2009, 11:21 AMdjmccabie
hi thanks for helping me. its actualyy (sinx - icosx) does this change things??

- Jan 28th 2009, 11:25 AMMoo