Question

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

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

Answer 1

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

Using
${\cos }^{-1}A-{\cos }^{-1}B={\cos }^{-1}(AB+\sqrt{(1-A^2)}\sqrt{(1-B^2)})$
$\therefore {\cos }^{-1}x-{\cos }^{-1}\frac{y}{2}=\alpha$
$\Rightarrow \frac{xy}{2}+\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}=\cos \alpha$
$\Rightarrow {(\cos \alpha -\frac{xy}{2})}^2=(1-x^2)(1-\frac{y^2}{2})$
$\Rightarrow 4x^2-4xy\cos \alpha +y^2=4(1-{\cos }^2\alpha )=4{\sin }^2\alpha$
Alternate Method:
${\cos }^{-1}x=\alpha +{\cos }^{-1}\frac{y}{2}\Rightarrow x=\cos (\alpha +{\cos }^{-1}\frac{y}{2})$
$\Rightarrow x=\frac{y}{2}\cos \alpha -\sin \alpha \sqrt{1-\frac{y^2}{4}}$
$\Rightarrow 2x-y\cos \alpha =-\sin \alpha \sqrt{1-\frac{y^2}{4}}$
$\Rightarrow {(2x-y\cos \alpha )}^2={\sin }^2\alpha (1-\frac{y^2}{4})$
$\Rightarrow 4x^2-4xy\cos \alpha +y^2=4(1-{\cos }^2\alpha )=4{\sin }^2\alpha$