Question ID: #1037
If $y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right)$ then $(x-y)^{2}+3y^{2}$ is equal to
Solution:
Given equation:
$$y = \cos\left(\frac{\pi}{3} + \cos^{-1}\left(\frac{x}{2}\right)\right)$$
Substitute $\frac{\pi}{3} = \cos^{-1}\left(\frac{1}{2}\right)$:
$$y = \cos\left(\cos^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{x}{2}\right)\right)$$
Using the trigonometric identity $\cos(\cos^{-1}A + \cos^{-1}B) = AB – \sqrt{1-A^2}\sqrt{1-B^2}$:
$$y = \left(\frac{1}{2}\right)\left(\frac{x}{2}\right) – \sqrt{1 – \left(\frac{1}{2}\right)^2}\sqrt{1 – \left(\frac{x}{2}\right)^2}$$
$$y = \frac{x}{4} – \sqrt{1 – \frac{1}{4}}\sqrt{1 – \frac{x^2}{4}}$$
$$y = \frac{x}{4} – \frac{\sqrt{3}}{2} \frac{\sqrt{4-x^2}}{2}$$
$$y = \frac{x – \sqrt{3}\sqrt{4-x^2}}{4}$$
Rearranging the terms:
$$4y = x – \sqrt{3}\sqrt{4-x^2}$$
$$x – 4y = \sqrt{3}\sqrt{4-x^2}$$
Squaring both sides:
$$(x – 4y)^2 = \left(\sqrt{3}\sqrt{4-x^2}\right)^2$$
$$x^2 – 8xy + 16y^2 = 3(4 – x^2)$$
$$x^2 – 8xy + 16y^2 = 12 – 3x^2$$
Bring all terms to one side:
$$4x^2 – 8xy + 16y^2 = 12$$
Divide the entire equation by $4$:
$$x^2 – 2xy + 4y^2 = 3$$
Split $4y^2$ into $y^2 + 3y^2$ to form a perfect square:
$$x^2 – 2xy + y^2 + 3y^2 = 3$$
$$(x – y)^2 + 3y^2 = 3$$
Ans. (3)
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