If x + frac{1}{x} = 2 cos theta, then x^3 + f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $x + \frac{1}{x} = 2 \cos 	heta$. We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Substituting $a = x$ and $b = \frac{

Vedclass · Accepted Answer

$2 \cos 3	heta$

Vedclass · Answer

(B) Given $x + \frac{1}{x} = 2 \cos 	heta$. We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$. Substituting $a = x$ and $b = \frac{1}{x}$,we get: $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}ight)^3 - 3 \left(x \cdot \frac{1}{x}ight) \left(x + \frac{1}{x}ight)$ $= (2 \cos 	heta)^3 - 3(1)(2 \cos 	heta)$ $= 8 \cos^3 	heta - 6 \cos 	heta$ $= 2(4 \cos^3 	heta - 3 \cos 	heta)$ Using the trigonometric identity $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta$,we get: $= 2 \cos 3	heta$.

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

Similar Questions

The value of $\cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{4 \pi}{7}\right) + \cos \left(\frac{6 \pi}{7}\right)$ is equal to

If $\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot \left(\frac{\pi}{24}\right)\right)=\sqrt{a^2+a}+\sqrt{a}$,then $a=$

If $L = \sin^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$ and $M = \cos^{2}\left(\frac{\pi}{16}\right) - \sin^{2}\left(\frac{\pi}{8}\right)$,then which of the following is correct?

If $90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2} =$

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module