1. ## complex exponential

Hi,
I need help using the complex exponential to express
cos6x as a polynomial in terms of cosx

I don't know how to get it in terms of cosx without sinx.

Thankyou,
ArTiCk

2. Originally Posted by ArTiCK
Hi,
I need help using the complex exponential to express
cos6x as a polynomial in terms of cosx

I don't know how to get it in terms of cosx without sinx.

Thankyou,
ArTiCk
$cos(6x) = Re[e^{6ix}] = Re[(cos(x) + i~sin(x))^6]$
where "Re" is the real part of the expression.

-Dan

3. Originally Posted by ArTiCK
Hi,
I need help using the complex exponential to express
cos6x as a polynomial in terms of cosx

I don't know how to get it in terms of cosx without sinx.

Thankyou,
ArTiCk
$\cos (6x) = Re\left[e^{i(6x)}\right] = Re \left[ (e^{ix})^6 \right]$

$= Re [ (\cos x + i \sin x)^6]$

$= Re[ \cos^6 x + 6i \cos^5 x \sin x - 15 \cos^4 x \sin^2 x - 20 i \cos^3 x \sin^3 x$ $+ 15 \cos^2 x \sin^4 x + 6i \cos x \sin^5 x - \sin^6 x]$

where the expansion is done using the binomial theorem

$= Re[(\cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x) + i ( .......)]$

$= \cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x$

$= \cos^6 x - 15 \cos^4 x (1 - \cos^2 x) + 15 \cos^2 x (1 - \cos^2 x)^2 - (1 - \cos^2 x)^3$

and I'll leave the expansion and subsequent simplification for you.

I get $18 \cos^6 x - 20 \cos^4 x + 4 \cos^2 x - 1$ ......

You should check that your answer gives 1 when you substitute x = 0. (Or that you get -1 when you substitute $x = \frac{\pi}{6}$ )

ps: I reserve the right for at least one error in this reply.