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

Accepted Answer

Answer: A. 4 sin2 \alpha {\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} ight)} ight) = \cos \alpha \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha \left( {\cos \alpha - {{xy} \over 2}} ight) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} squaring both sides {x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha 	herefore 4x2 – 4xy cos \alpha + y2 = 4 {\sin ^2}\alpha

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 :

Answer

Explanation

More JEE Math questions