1) Use De Moivre’s theorem to show that $\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} cis (5x)$

2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x).

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- Oct 25th 2006, 08:56 PMscorpion007complex number question
1) Use De Moivre’s theorem to show that $\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} cis (5x)$

2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x). - Oct 25th 2006, 09:01 PMscorpion007
ah i got the first one... i think.

- Oct 25th 2006, 09:52 PMCaptainBlack
This should be:

$\displaystyle

(1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} [\cos (5x) + i \sin(5x)]

$

Quote:

2) Hence, find expressions for cos (5x) and sin (5x) in terms of tan(x) and cos(x).

$\displaystyle (1 + i \tan(x))^5 = \frac{1}{\cos^5(x)} (\cos (5x) + i \sin(5x))$

implies:

$\displaystyle \cos^5(x) (1 + i \tan(x))^5 = \cos (5x) + i \sin(5x)$

so expanding the second term on the left and equating real and imaginary

parts we will have the answer required - but I have to rush off now, may be

able to finish this later.

RonL - Oct 25th 2006, 10:14 PMscorpion007
hi, CaptainBlack, thanks. For the actual equation i wrote "cis" which is not a typo, as you guessed "cis(x)" means "cos(x + i sin(x)". It is just an abbreviated form we commonly use. Maybe you don't use it where you are.

Now for the second part, is there a fairly simple way to expand (1 + i tan(x))^5 ? Expanding a degree 5 poly is a bit of a pain isn't it? - Oct 25th 2006, 11:20 PMCaptainBlack
(1 + i tan(x))^5 = 1 + 5 (i tan(x)) + 10 (i tan(x))^2 + 10(i tan(x))^3 +

.........................5(i tan(x))^4 + (i tan(x))^5

.......................= 1 + 5 i tan(x) - 10 (tan(x))^2 - 10 i (tan(x))^3 +

.........................5(tan(x))^4 + i (tan(x))^5

.......................= [1 - 10 (tan(x))^2 + 5(tan(x))^4] +

..........................i [5 tan(x) - 10 (tan(x))^3 + (tan(x))^5]

RonL