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
$\displaystyle \cos (6x) = Re\left[e^{i(6x)}\right] = Re \left[ (e^{ix})^6 \right]$
$\displaystyle = Re [ (\cos x + i \sin x)^6]$
$\displaystyle = Re[ \cos^6 x + 6i \cos^5 x \sin x - 15 \cos^4 x \sin^2 x - 20 i \cos^3 x \sin^3 x$ $\displaystyle + 15 \cos^2 x \sin^4 x + 6i \cos x \sin^5 x - \sin^6 x]$
where the expansion is done using the binomial theorem
$\displaystyle = Re[(\cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x) + i ( .......)]$
$\displaystyle = \cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x$
$\displaystyle = \cos^6 x - 15 \cos^4 x (1 - \cos^2 x) + 15 \cos^2 x (1 - \cos^2 x)^2 - (1 - \cos^2 x)^3$
and I'll leave the expansion and subsequent simplification for you.
I get $\displaystyle 18 \cos^6 x - 20 \cos^4 x + 4 \cos^2 x - 1$ ......
You should check that your answer gives 1 when you substitute x = 0. (Or that you get -1 when you substitute $\displaystyle x = \frac{\pi}{6}$ )
ps: I reserve the right for at least one error in this reply.