Question

If $co{s}^{-1}x-co{s}^{-1}\left(\frac{y}{2}\right)=\alpha$, then $4x^2-4xy\cos \alpha +y^2=$

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

Answer 1

Answer: (B)

$4{\sin }^2\alpha$

Consider the given equation.

${\cos }^{-1}x-{\cos }^{-1}\left(\frac{y}{2}\right)=\alpha$ …….. (1)

We know that

${\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}\{xy+\sqrt{1-x^2}\sqrt{1-y^2}\}$

Therefore,

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

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

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

On squaring both sides, we get

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

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

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

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

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

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

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

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

Hence, this is the answer.