$$\sin6\alpha\equiv \sin2\alpha(16\cos^4\alpha-16\cos^2\alpha+3)$$

Can you help me with De Moivre's theorem and how I would go about tackling this question.

I understand that De Moivre's theorem states that $(\cos\alpha+i\sin\alpha)^n \equiv \cos n\alpha+i\sin n\alpha$.

But I don't see how this would come to use in this question.


From De Moivres $$ \cos 6\alpha+i\sin 6\alpha=(\cos\alpha+i\sin \alpha)^6=\\(\cos^6\alpha+\ldots)+i\left({6\choose 1}\cos^5\alpha\sin\alpha-{6\choose 3}\cos^3\alpha\sin^3\alpha+{6\choose 5}\cos\alpha\sin^5\alpha\right). $$ Therefore $$\sin 6\alpha={6\choose 1}\cos^5\alpha\sin\alpha+{6\choose 3}\cos^3\alpha\sin^3\alpha+{6\choose 5}\cos\alpha\sin^5\alpha=\\ 2\cos\alpha\sin\alpha(\ldots)=\ldots$$

  • $\begingroup$ why do you only do the 2nd, 4th and 6th term, of the binomial series? (even terms. And then with cos, the odd terms) $\endgroup$ – maxmitch May 31 '13 at 11:39
  • 1
    $\begingroup$ @maxmitch: Because $i^1=i,i^3=-i,i^5=i$ and $i^0=1,i^2=-1,i^4=1,i^6=-1$. $\endgroup$ – P.. May 31 '13 at 11:54
  • $\begingroup$ Of course!! My bad. $\endgroup$ – maxmitch May 31 '13 at 13:36


$$\sin 6\alpha=\sin (3\cdot 2\alpha)=3\sin2\alpha-4\sin^32\alpha=\sin2\alpha(3-4\sin^22\alpha)$$

Now, $$3-4\sin^22\alpha=3-4(2\sin\alpha\cos\alpha)^2=3-16\cos^2\alpha(1-\cos^2\alpha)$$

  • $\begingroup$ @maxmitch, how about this? $\endgroup$ – lab bhattacharjee May 31 '13 at 12:36

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