Question

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

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

Answer 1

Answer: (C)

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