Question

If $2\cos \theta =x+\frac{1}{x},$ then $2$ $\cos 3\theta$ is equal to

• A. $x^3+\frac{1}{x^3}$
• B. $x^2+\frac{1}{x^2}$
• C. $\frac{x^2}{1+x^2}$
• D. none of these

Answer 1

Answer: (A)

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

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

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

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

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

$=(x+\frac{1}{x})[{(x+\frac{1}{x})}^2-3]$

$=(x+\frac{1}{x})[x^2+\frac{1}{x^2}+2-3]$

$=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)$

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

$2\cos 3\theta =x^3+\frac{1}{x^3}$.