If [ ^ - 1x - ^ - 1y 2 = ],where --1 [ ] x [ ] 1, -- 2 [ ] y [ ] 2, x [ ] [y 2] , then for…

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 :

Accepted Answer

$${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$

$$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $$

$$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}}  = \cos \alpha $$

$$\left( {\cos \alpha  - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} $$

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 $${\sin ^2}\alpha $$

Solutions · 1