If cos^{-1} x - cos^{-1} frac{y}{2} = alpha,w | vedclass

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(D) Given $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$. Taking $\cos$ on both sides,we get $\cos(\cos^{-1} x - \cos^{-1} \frac{y}{2}) = \cos \alpha$

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$4 \sin^2 \alpha$

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(D) Given $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$. Taking $\cos$ on both sides,we get $\cos(\cos^{-1} x - \cos^{-1} \frac{y}{2}) = \cos \alpha$. Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we have: $x \cdot \frac{y}{2} + \sqrt{1 - x^2} \sqrt{1 - \frac{y^2}{4}} = \cos \alpha$. Rearranging the terms: $\sqrt{1 - x^2} \sqrt{1 - \frac{y^2}{4}} = \cos \alpha - \frac{xy}{2}$. Squaring both sides: $(1 - x^2)(1 - \frac{y^2}{4}) = (\cos \alpha - \frac{xy}{2})^2$. $1 - \frac{y^2}{4} - x^2 + \frac{x^2y^2}{4} = \cos^2 \alpha - xy \cos \alpha + \frac{x^2y^2}{4}$. $1 - \frac{y^2}{4} - x^2 = \cos^2 \alpha - xy \cos \alpha$. Multiply the entire equation by $4$: $4 - y^2 - 4x^2 = 4 \cos^2 \alpha - 4xy \cos \alpha$. Rearranging to find the value of $4x^2 - 4xy \cos \alpha + y^2$: $4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos^2 \alpha$. Since $1 - \cos^2 \alpha = \sin^2 \alpha$,we get: $4x^2 - 4xy \cos \alpha + y^2 = 4 \sin^2 \alpha$.

If $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$,where $-1 \le x \le 1$,$-2 \le y \le 2$,and $x \le \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

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