STD 11 - 3. trignometrical ratios,functions and identities — Maths STD 11 Science — Question

On squaring both sides,

we get $x + \frac{1}{x} + 2 = 4\,{\cos ^2}\theta $

==> $x + \frac{1}{x} = 4{\cos ^2}\theta - 2$

==> $x + \frac{1}{x} = $$2(2{\cos ^2}\theta - 1)$ $ = 2\cos 2\theta $…..$(ii)$

Again squaring both sides, ${x^2} + \frac{1}{{{x^2}}} + 2 = 4{\cos ^2}2\theta $

==> ${x^2} + \frac{1}{{{x^2}}} = 4{\cos ^2}2\theta - 2$$ = 2(2{\cos ^2}2\theta - 1)$

==> ${x^2} + \frac{1}{{{x^2}}} = 2\cos 4\theta $…..$(iii)$

Now take cube of both sides, ${\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^3} = {(2\cos 4\theta )^3}$

==> ${x^6} + \frac{1}{{{x^6}}} + 3{x^2} \times \frac{1}{{{x^2}}}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 8{\cos ^3}4\theta $

==> ${x^6} + \frac{1}{{{x^6}}} + 3\,(2\cos 4\theta ) = 8{\cos ^3}4\theta $

$ \Rightarrow {x^6} + \frac{1}{{{x^6}}} = 8{\cos ^3}4\theta - 6\cos 4\theta $

$= 2\,(4{\cos ^3}4\theta - 3\cos 4\theta )$

$= 2\cos 3(4\theta )  =  2\cos 12\theta $.