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} - 4 x^{2} + x^{2} y^{2}}}{2}\right) = \alpha$
$ \Rightarrow 4 - y^{2} - 4 x^{2} + x^{2}y^{2} $
$= 4 \cos^{2} \alpha + x^{2} y^{2} - 4 xy \cos\alpha $
$\Rightarrow 4x^{2} + y^{2} - 4xy \cos \alpha = 4 \sin^{2} \alpha. $