If ${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $,where –1 $ \le $ x $ \le $ 1, – 2 $ \le $ y $ \le $ 2, x $ \le $ ${y \over 2}$ , then for all x, y, 4x2 – 4xy cos $\alpha $ + y2 is equal to :

Question

If ${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $,where –1 $ \le $ x $ \le $ 1, – 2 $ \le $ y $ \le $ 2, x $ \le $ ${y \over 2}$ , then for all x, y, 4x² – 4xy cos $\alpha $ + y² is equal to :

4 sin² $\alpha $

2 sin² $\alpha $

4 sin² $\alpha $ - 2x²y²

4 cos² $\alpha $ + 2x²y²

Solution

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