# Thread: De moivres theorem

1. ## 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?

2. Hello,
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 : $\displaystyle (\cos(x)+i\sin(x))^n=\cos(nx)+i \sin(nx)$

Remember that the sine function is odd and the cosine function is even.
So $\displaystyle (\cos(x)-i\sin(x))^n=(\cos(-x)+i \sin(-x))^n=$ $\displaystyle \cos(-nx)+i \sin(-nx)=\cos(nx)-i \sin(nx)$

3. so did i do it right :S

is it different when cosine and sine are switched round?

thanks

4. 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 $\displaystyle (-\cos(x)+i \sin(x))^n$ ?
Then it's just $\displaystyle (-1)^n (\cos(x)-i\sin(x))^n=(-1)^n(\cos(nx)-i\sin(nx))$

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

6. Originally Posted by djmccabie
hi thanks for helping me. its actualyy (sinx - icosx) does this change things??
Yes :

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

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

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

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# express (cosx isinx/sinx icosx)5 in the form cosA isinA solution

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