Question

If $2\sin \theta =x^2+\frac{1}{x^2}$ then $x^6+\frac{1}{x^6}$ is equal to

• A. $2\sin \theta (1-2\cos \theta )(1+2\cos \theta )$
• B. $2\cos \theta (1-2\cos \theta )(1+2\cos \theta )$
• C. $2\sin \theta (1-2\sin \theta )(1+2\sin \theta )$
• D. $2\sin \theta (2\sin \theta -\sqrt{3})(2\sin \theta +\sqrt{3})$

Answer 1

The correct option is D $2\sin \theta (2\sin \theta -\sqrt{3})(2\sin \theta +\sqrt{3})$

$x^2+\frac{1}{x^2}=2\sin \theta$

$x^6+\frac{1}{x^6}={\left(x^2\right)}^3+\frac{1}{{\left(x^2\right)}^3}$ is of the form $a^3+b^3=(a+b)(a^2=b^2-ab)$

$=(x^2+\frac{1}{x^2})(x^4+\frac{1}{x^4}-x^2\times \frac{1}{x^2})$

$=\left(2\sin \theta \right)(x^4+\frac{1}{x^4}-1)$

$=\left(2\sin \theta \right)({(x^2+\frac{1}{x^2})}^2-2x^2\times \frac{1}{x^2}-1)$

$=\left(2\sin \theta \right)({\left(2\sin \theta \right)}^2-2-1)$

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

$=2\sin \theta \cdot ({\left(2\sin \theta \right)}^2-{\left(\sqrt{3}\right)}^2)$

$=2\sin \theta (2\sin \theta -\sqrt{3})(2\sin \theta +\sqrt{3})$