Question

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

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

Answer 1

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

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