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Question

(A) Given the quadratic equation $x^2 + x \sin 	heta - 2 \sin 	heta = 0$.By Vieta's formulas,the sum of roots is $\alpha + \beta = -\sin 	heta$ and

Vedclass · Accepted Answer

$\frac{2^{12}}{(\sin 	heta + 8)^{12}}$

Vedclass · Answer

(A) Given the quadratic equation $x^2 + x \sin 	heta - 2 \sin 	heta = 0$.By Vieta's formulas,the sum of roots is $\alpha + \beta = -\sin 	heta$ and the product of roots is $\alpha \beta = -2 \sin 	heta$.We need to evaluate the expression $E = \frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$.Simplifying the denominator: $\alpha^{-12} + \beta^{-12} = \frac{1}{\alpha^{12}} + \frac{1}{\beta^{12}} = \frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}}$.Substituting this into the expression: $E = \frac{\alpha^{12} + \beta^{12}}{\left(\frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}}ight)(\alpha - \beta)^{24}} = \frac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}}$.Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4 \alpha \beta$,we have $(\alpha - \beta)^{24} = ((\alpha + \beta)^2 - 4 \alpha \beta)^{12}$.Thus,$E = \left[ \frac{\alpha \beta}{(\alpha + \beta)^2 - 4 \alpha \beta} ight]^{12}$.Substituting the values of $\alpha + \beta$ and $\alpha \beta$: $E = \left[ \frac{-2 \sin 	heta}{(-\sin 	heta)^2 - 4(-2 \sin 	heta)} ight]^{12} = \left[ \frac{-2 \sin 	heta}{\sin^2 	heta + 8 \sin 	heta} ight]^{12}$.Since $	heta \in (0, \frac{\pi}{2})$,$\sin 	heta 
eq 0$,we can cancel $\sin 	heta$: $E = \left( \frac{-2}{\sin 	heta + 8} ight)^{12} = \frac{2^{12}}{(\sin 	heta + 8)^{12}}$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + x \sin \theta - 2 \sin \theta = 0$,where $\theta \in (0, \frac{\pi}{2})$,then the value of $\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to

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