Question

If $\cos x^{-1}-{\cos }^{-1}\left(\frac{y}{2}\right)=a$, then $4x^2-4xy\cos x+y^2$ is equal to

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

Answer 1

Answer: (C)

$4{\sin }^{2}a$

Let ${\cos }^{-1}x=\alpha$ & ${\cos }^{-1}\frac{y}{2}=\beta$

$\alpha -\beta =a$

$\cos (\alpha -\beta )=\cos a$

$\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos a$

$x\frac{y}{2}-\sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}=\cos a$

$\frac{xy}{2}-\sqrt{1-x^2}\frac{\sqrt{4-y^2}}{2}=\cos a$

$\frac{xy}{2}-\cos a=\frac{\sqrt{1-x^2}\sqrt{4-y^2}}{2}$

On squaring, we have

$\Rightarrow \frac{x^2{y}^2}{4}+{\cos }^{2}a-xy\cos a=\frac{(1-x^2)(4-y^2)}{4}$

$\Rightarrow \frac{x^2{y}^2}{4}-xy\cos a=\frac{(1-x^2)(4-y^2)}{4}-{\cos }^{2}a$

$\Rightarrow x^2{y}^2-4xy\cos a=4-y^2-4x^2+x^2{y}^2-4{\cos }^{2}a$

$\Rightarrow 4x^2-4xy\cos a+y^2=4-4{\cos }^{2}a$

$\Rightarrow 4x^2-4xy\cos a+y^2=4(1-{\cos }^{2}a)$

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