If and are the roots of x^2+x-2=0, [0,2], then ^12+^12(^-12+^-12)(-)^24 is equal to:

Question

If $\alpha$ and $\beta$ are the roots of $x^2+x\sin	heta-2\sin	heta=0$, $	heta\in\left[0,\dfrac\pi2ight]$, then $\dfrac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha-\beta)^{24}}$ is equal to:

Accepted Answer

$\alpha+\beta=-\sin\theta$, $\alpha\beta=-2\sin\theta$.

$$\frac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha-\beta)^{24}}=\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}.$$

$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\sin^2\theta+8\sin\theta$, so

$$\frac{(2\sin\theta)^{12}}{\sin^{12}\theta(\sin\theta+8)^{12}}=\frac{2^{12}}{(\sin\theta+8)^{12}}.$$

Solutions · 1