If x+frac{1}{x}=2 sin alpha and y+frac{1}{y}= | vedclass

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(C) Given $x+\frac{1}{x}=2 \sin \alpha$. Solving for $x$ using the quadratic formula $x^2 - (2 \sin \alpha)x + 1 = 0$,we get $x = \sin \alpha \pm i \c

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$2 \sin 3(\beta-\alpha)$

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(C) Given $x+\frac{1}{x}=2 \sin \alpha$. Solving for $x$ using the quadratic formula $x^2 - (2 \sin \alpha)x + 1 = 0$,we get $x = \sin \alpha \pm i \cos \alpha = \cos(\frac{\pi}{2} - \alpha) \pm i \sin(\frac{\pi}{2} - \alpha) = e^{\pm i(\frac{\pi}{2} - \alpha)}$.Taking $x = e^{i(\frac{\pi}{2} - \alpha)}$,then $x^3 = e^{i(\frac{3\pi}{2} - 3\alpha)}$.Given $y+\frac{1}{y}=2 \cos \beta$. Solving for $y$,we get $y = \cos \beta \pm i \sin \beta = e^{\pm i\beta}$.Taking $y = e^{i\beta}$,then $y^3 = e^{i3\beta}$.Thus,$x^3 y^3 = e^{i(\frac{3\pi}{2} - 3\alpha + 3\beta)} = \cos(\frac{3\pi}{2} + 3(\beta - \alpha)) + i \sin(\frac{3\pi}{2} + 3(\beta - \alpha)) = \sin(3(\beta - \alpha)) - i \cos(3(\beta - \alpha))$.Similarly,$\frac{1}{x^3 y^3} = \sin(3(\beta - \alpha)) + i \cos(3(\beta - \alpha))$.Adding these,$x^3 y^3 + \frac{1}{x^3 y^3} = 2 \sin(3(\beta - \alpha))$.Therefore,the correct option is $C$.

If $x+\frac{1}{x}=2 \sin \alpha$ and $y+\frac{1}{y}=2 \cos \beta$,then $x^3 y^3+\frac{1}{x^3 y^3}=$

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