JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 15)
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
Explanation
$${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$
$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $$
$$\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} $$
squaring both sides
$${x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha $$
$$ \therefore $$ 4x2 – 4xy cos $$\alpha $$ + y2 = 4 $${\sin ^2}\alpha $$
$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$
$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $$
$$\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} $$
squaring both sides
$${x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha $$
$$ \therefore $$ 4x2 – 4xy cos $$\alpha $$ + y2 = 4 $${\sin ^2}\alpha $$
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