How to use de moivre's theorem to express #(sin 6theta)/sin theta# as a polynomial in #cos theta#?

1 Answer
George C.
Feb 5, 2018

#(sin 6 theta)/sin theta = 32cos^5 theta-32cos^3 theta+6cos theta#

Explanation:

For brevity write #c# for #cos theta# and #s# for #sin theta#

By Pythagoras' theorem, we have:

#c^2+s^2 = 1#

and hence:

#s^2=1-c^2#

By de Moivre's theorem, we have:

#cos 6 theta + i sin 6 theta#

#=(c+is)^6#

#=c^6+6ic^5s-15c^4s^2-20ic^3s^3+15c^2s^4+6ics^5-s^6#

#=(c^6-15c^4s^2+15c^2s^4-s^6)+is(6c^5-20c^3s^2+6cs^4)#

Equating imaginary parts, we have:

#sin 6 theta = s(6c^5-20c^3s^2+6cs^4)#

#color(white)(sin 6 theta) = s(6c^5-20c^3(1-c^2)+6c(1-c^2)^2)#

#color(white)(sin 6 theta) = s(6c^5-20c^3+20c^5+6c-12c^3+6c^5)#

#color(white)(sin 6 theta) = s(32c^5-32c^3+6c)#

So:

#(sin 6 theta)/sin theta = 32cos^5 theta-32cos^3 theta+6cos theta#

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