$$ \text { If } y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \text {, then }(x-y)^2+3 y^2 \text { is equal to } $$
Answer (integer)
3
Solution
<p>$$\begin{aligned}
& y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\
& =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\
& =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\
& \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\
& \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\
& \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\
& x^2+4 y^2-2 x y=3 \\
& (x-y)^2+3 y^2=3
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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