If y = cos left(frac{pi}{3} + cos^{-1} frac{x | vedclass

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(C) Given $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$.Since $\cos^{-1} \frac{1}{2} = \frac{\pi}{3}$,we have $y = \cos \left(\cos^{-1

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$3$

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(C) Given $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$.Since $\cos^{-1} \frac{1}{2} = \frac{\pi}{3}$,we have $y = \cos \left(\cos^{-1} \frac{1}{2} + \cos^{-1} \frac{x}{2}\right)$.Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:$y = \left(\frac{1}{2}\right) \left(\frac{x}{2}\right) - \sqrt{1 - \left(\frac{1}{2}\right)^2} \sqrt{1 - \left(\frac{x}{2}\right)^2}$.$y = \frac{x}{4} - \sqrt{\frac{3}{4}} \sqrt{\frac{4 - x^2}{4}} = \frac{x - \sqrt{3(4 - x^2)}}{4}$.$4y = x - \sqrt{3(4 - x^2)}$.Rearranging and squaring both sides:$(4y - x)^2 = 3(4 - x^2)$.$16y^2 + x^2 - 8xy = 12 - 3x^2$.$4x^2 + 16y^2 - 8xy = 12$.Dividing by $4$:$x^2 + 4y^2 - 2xy = 3$.We can rewrite this as $x^2 - 2xy + y^2 + 3y^2 = 3$.$(x - y)^2 + 3y^2 = 3$.

If $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$,then $(x - y)^2 + 3y^2$ is equal to . . . . . . .

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