Question

Prove that ${\cos }^3\theta +{\cos }^3(\theta -\frac{2\pi }{n})+{\cos }^3(\theta -\frac{4\pi }{n})+...$ to $n$ terms $=0$

Answer 1

As we know that

$\cos 3x=4{\cos }^{3}x-3\cos x{\cos }^{3}x=\frac{3\cos x+\cos 3x}{4}$

Using that formula, we can write

${\cos }^3\theta +{\cos }^3(\theta -\frac{2\pi }{n})+{\cos }^3(\theta -\frac{4\pi }{n})+...tonterms=\frac{3[\cos \theta +\cos (\theta -\frac{2\pi }{n})+\cos (\theta -\frac{4\pi }{n})+...tonterms]+[\cos 3\theta +\cos (3\theta -\frac{6\pi }{n})+\cos (3\theta -\frac{12\pi }{n})+...tonterms]}{4}$

As we seen from the above fromula that if $\beta =\frac{2\pi N}{n}whereN=1,2,\mathrm{3....}$

So, the sum of both of series is equal to zero.

Hence, ${\cos }^3\theta +{\cos }^3(\theta -\frac{2\pi }{n})+{\cos }^3(\theta -\frac{4\pi }{n})+...tonterms=0$