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

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

Question

(B) Given that $x + \frac{1}{x} = 2\cos 	heta$.We use 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 that $x + \frac{1}{x} = 2\cos 	heta$.We use 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 obtain:$= 2\cos 3	heta$.

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

Similar Questions

The value of $\sin \frac{\pi }{14} \sin \frac{3\pi }{14} \sin \frac{5\pi }{14} \sin \frac{7\pi }{14} \sin \frac{9\pi }{14} \sin \frac{11\pi }{14} \sin \frac{13\pi }{14}$ is equal to

If $A = \sin^8 \theta + \cos^{14} \theta$,then for all real values of $\theta$:

$\Delta XYZ$ is right-angled at $Y$. If $m \angle X = 60^{\circ}$,then find the value of $\left(\sec Z + \frac{2}{\sqrt{3}}\right)$.

Let $A = \{ \theta : 2\cos^2 \theta + \sin \theta \le 2 \}$ and $B = \{ \theta : \frac{\pi}{2} \le \theta \le \frac{3\pi}{2} \}$,then $A \cap B$ is

$\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module