Question

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

• A. $2{\sin }^2\alpha$
• B. $4{\sin }^2\alpha -2x^2{y}^2$
• C. $4{\sin }^2\alpha$
• D. $4{\cos }^2\alpha +2x^2{y}^2$

Answer 1

The correct option is C $4{\sin }^2\alpha$

${\cos }^{-1}x-{\cos }^{-1}\frac{y}{2}=\alpha$
$\Rightarrow \cos ({\cos }^{-1}x-{\cos }^{-1}\frac{y}{2})=\cos \alpha$
$[\because {\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}(xy+\sqrt{1-x^2}\sqrt{1-y^2})]$

$\Rightarrow \frac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}=\cos \alpha$
$\Rightarrow (\cos \alpha -\frac{xy}{2})=\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}$
Squaring both sides
$\Rightarrow {\cos }^2\alpha +\frac{x^2{y}^2}{4}-xy\cos \alpha =(1-x^2)(1-\frac{y^2}{4})$
$\Rightarrow x^2-xy\cos \alpha +\frac{y^2}{4}=1-{\cos }^2\alpha ={\sin }^2\alpha$
$\Rightarrow 4x^2-4xy\cos \alpha +y^2=4{\sin }^2\alpha$