Question

If $2\cos \theta =x+\frac{1}{x}$, then prove that $2\cos 3\theta =x^3+\frac{1}{x^2}$

Answer 1

Formula:

$a^3+b^3=(a+b{)}^3-3ab(a+b)$

Given that,

$2\cos \theta =x+\frac{1}{x}$

$⟹\cos \theta =\frac{1}{2}(x+\frac{1}{x})$

$2\cos 3\theta =2(4{\cos }^3\theta -3\cos \theta )=8{\cos }^3\theta -6cos\theta$

$=8.\frac{1}{8}(x+\frac{1}{x}{)}^3-6.\frac{1}{2}(x+\frac{1}{x})$

$=(x+\frac{1}{x}{)}^3-3(x+\frac{1}{x})$

$=(x+\frac{1}{x}{)}^3-3.x.\frac{1}{x}(x+\frac{1}{x})$

$=x^3+\frac{1}{x^3}$