Question

If $co{s}^{-1}x-co{s}^{-1}\left(y/2\right)=\alpha$, then $400(4x^2-4xycos\alpha +y^2)+328si{n}^2\alpha$ is equal to $asi{n}^2\alpha$ where $a=$

Answer 1

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

$⟹x=\cos \alpha \left(y/2\right)-\sin \alpha \sqrt{1-y^2/4}$

$⟹(2x-y\cos \alpha {)}^2=(4-y^2){\sin }^2\alpha$

$⟹4x^2-4xy\cos \alpha +y^2=4{\sin }^2\alpha$

$⟹400\left(4{\sin }^2\alpha \right)+328{\sin }^2\alpha$

$⟹1928{\sin }^2\alpha$

$⟹a=1928$