If$\ \cos{ heta = \frac{1}{2}\left( x + \frac{1}{x} ight)},\ then\ \frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} ight)$ is equal to

Question

If$\ \cos{	heta = \frac{1}{2}\left( x + \frac{1}{x} ight)},\ then\ \frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} ight)$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given that, $\cos{	heta = \frac{1}{2}\left( x + \frac{1}{x} ight)}$
$\Rightarrow x + \frac{1}{x} = 2\cos	heta$ ….(i)
We know that, $x^{2} + \frac{1}{x^{2}} = \left( x + \frac{1}{x} ight)^{2} - 2$
$= \left( 2\cos	heta ight)^{2} - 2 = 4\cos^{2}{	heta - 2}$
$= 2\cos{2	heta}\ \ \$[from Eq.(i)]
$	herefore\ \frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} ight) = \frac{1}{2} 	imes 2\cos{2	heta = \cos{2	heta}}$

Q. If$\ \cos{\theta = \frac{1}{2}\left( x + \frac{1}{x} \right)},\ then\ \frac{1}{2}\left( x^{2} + \frac{1}{x^{2}} \right)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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