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. $2sin2\alpha$
• C. $-4sin^2\alpha$
• D. $4sin^2\alpha$

Answer 1

Answer: (D)

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